InterviewSolution
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InterviewSolution
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`N_(2)O_(4)` dissociates as `N_(2)O_(4)(g)hArr2NO_(2)(g)` At `40^(@)C` and one atmosphere `%` decomposition of `N_(2)O_(4)` is `50.3%`. At what pressure and same temperature, the equilibrium mixture has the ratio of `N_(2)O_(4): NO_(2)` as `1:8`? |
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Answer» Case I `N_(2)O_(4)(g)hArr2NO_(2)(g)` `{:("At Eq",,(1-x),,2x,),(,,"total moles" =1+x,,):}` `p_(N_(2)O_(4))=((1-x))/((1+x))xxP, p_(NO_(2))=(2x)/((1+x))xxP` `K_(p)=(((2x)/(1+x).P)^(2))/(((1-x)/(1+x).P))=(4x^(2)P)/((1-x^(2)))` The `%` dissociation of `N_(2)O_(4)=50.3%` (given) Hence, degree of dissociation `=50.3/100=0.503` and `P=1` `:. K_(P)=(4xx(0.503)^(2)xx1)/([1-(0.503)^(2)])` `rArr K_(p)=1.3548` atm Case II `N_(2)O_(4)hArr2NO_(2)` `{:((1-x),(2x)):}` Given, `((1-x))/(2x)=1/8` `x=0.8` Let the new pressure be `P` atm. `K_(p)=(4x^(2)P)/((1-x^(2)))=(4xx0.8xx0.8xxP)/((1+0.8)(1-0.8))=1.3548` `P=0.19 "atm"` |
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