InterviewSolution
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InterviewSolution
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The density of an equilibrium mixture of `N_(2)O_(4)` and `NO_(2)` at `1` atm is `3.62 g L^(-1)` at `288 K` and `1.84 g L^(-1)` at `348 K`. Calculate the entropy change during the reaction at `348 K`. |
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Answer» `N_(2)O_(4) hArr 2NO_(2)` For `K_(p)`, proceed as follows: `PV=nRT=w/m_(mix) RT` `rArr m_(mix)=w/Vxx(RT)/(P)=(dRT)/(P)` `=3.62xx0.082xx288=85.6` Let a mol of `N_(2)O_(4)` and `(1-a)` mol of `NO_(2)` exist at equilibrium. `:. axx92+(1-a)xx46=85.6` `:. a=0.86` `:. n_(N_(2)O_(4))=0.86, n_(NO_(2))=0.14 "mol"` `K_(p)=(0.14xx0.14)/(0.86)xx[1/1]^(1)=0.0228` atm at `288 K` Case II `m_(mix)=(dRT)/(P)=1.84xx0.0821xx348=52.57` Let a mol of `N_(2)O_(4)` and `(1-a)` mol of `NO_(2)` exist at equilibrium. `:. axx92+(1-a)xx46=52.57` `:. a=0.14` `:. n_(N_(2)O_(4))=0.14, n_(NO_(2))=0.86` `:. K_(p)=(0.86xx0.86)/(0.14)[1/1]^(1)=5.283` "atm" at `348 K` `log_(10)=(K_(P_(2))/(K_(P_(1))))=(DeltaH)/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` `rArr "log"_(10) 5.283/0.0228=(DeltaH)/(2xx2.303)[(348-288)/(348xx288)]` `:. DeltaH=181956 cal=18.196` kcal `DeltaG=-2.303 RT log K_(p)` `=-2.303xx2xx348xxlog 5.283` `=-1158.7 cal` `DeltaS=(DeltaH-DeltaG)/(T)=(18195.6+1158.7)/(348)=55.62 cal` |
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