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For the reaction N_(2)+2O_(2)to2NO_(2) Given : at 1 atm, 300 K S_(N_(2))=180J//mol//K C_(p)(N_(2))=30J//mol//K S_(O_(2))=220J//mol//K C_(p)(O_(2))=30J//mol//K S_(NO_(2))=240J//mol//K C_(p)(NO_(2))=40J//mol//K Calculate (i) DeltaS_(300K,1"atm") (ii) DeltaS_(400K,1"atm") (iii) DeltaS_(300K,5"atm") (iv) DeltaS_(400K,5"atm") |
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Answer» Solution :(i) `(DeltaS_(R))_(300)=2S_(NO_(2))-2S_(O_(2))-S_(N_(2))` `=2xx240-2xx220-180` = `-140Jmol^(-1)K^(-1)` `(DeltaCp)_(r)=2Cp(NO_(2))-2Cp(O_(2))-Cp(N_(2))` = `2xx40-2xx30-30` `=-10Jmol^(-1)k^(-1)` (ii) `(DeltaS_(r))_(400)=(DeltaS_(r))_(300)+(DeltaC_(p))_(r)//nT_(2)/T_(1)` = `-140-10//n4/3` `=-142.88Jmol^(-1)k^(-1)` (III) `(DeltaS_(r))_(300k,5"atm")=(DeltaS_(r))_(300k,1"atm")+Deltan_(g)R//n` `p_(1)/p_(2)=-140+(-1)R//n1/5=-140+R//n5` = `-140+8.314//n5` = `-126.62 J mol^(-1)k^(-1)` (iv) `(DeltaS_(r))_(400k,5"atm")=(DeltaS_(r))_(400k,"1atm")-R//n1/5` = `142.88+R//n5` = `-129.5Jmol^(-1)k^(-1)` |
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