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The density of an equilibrium mixture of...

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`.

Text Solution

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`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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