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`N_(2)``O_(4)` is `25%` dissociated at `37^(@)C` and one atmosphere pressure. Calculate (i) `K_(p)` and (ii) the percentage dissociation at 0.1 atm and `37^(@)C` |
Answer» Correct Answer - (i) `0.266` atm (ii) `63.25%` `N_(2)O_(4)hArr2NO_(2)` `{:(1,0,,,),(1-.25,.50,,,),(.75,.50,,,n_("total")=1.25):}` `P_(N_(2)O_(4))=((P_(NO_(2)))^(2))/((P_(N_(2)O_(4))))=(.50//1.25)^(2)/((.75//1.25))=(.50xx.50)/(1.25xx.75)=(4)/(15)=0.266` At pressure `0.1` atm `N_(2)O_(4)hArr2NO_(2)` `1 0 ` `(1-alpha) 2a` `K_(P)=((1-alpha)/(1+alpha))xx0.1 , P_(NO_(2)=(((2alpha))/(1+alpha))xx0.1` `K_(P)=((2alpha)/(1+alpha)xx0.1)^(2)/(((1-alpha)/(1+alpha))xx0.1), K_(P)=(4alpha^(2)xx0.1)/((1+alpha)(1-alpha))` `0.266=(0.1xx4alpha^(2))/(1-alpha^(2))` `0.665=(1+0.665)alpha^(2). rArr alpha=63.25%` |
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