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InterviewSolution
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The `K_(p)` for the reaction `N_(2)O_(4)hArr2NO_(2)` is `640 mm` at `775 K`. Calculate the percentage dissociation of `N_(2)O_(4)` at equilibrium pressure of `160 mm`. At what pressure, the dissociation will be `50%`? |
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Answer» `{:(,N_(2)O_(4),hArr,2NO_(2)),("Mole before equilibrium",1,,0),("Mole at equilibrium",(1-x),,2x):}` `K_(p)=(4x^(2))/((1-x))xx[(P)/(sumn)]^(Deltan)` `640=(4x^(2))/((1-x))(160)/((1+x))` `4=(4x^(2))/(1-x^(2))` or `1-x^(2)=x^(2)` or `2x^(2)=1` ` :. x^(2)=1//2` or `x=0.707=70.7%` Also `640=(4xx(0.5)^(2))/(0.5)xx(P)/(1.5)` (if `x=0.5`) `:. P=480 mm` |
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