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You are given normal boiling points and standard enthalpies of vaporization, Calendly. the entropy of vaporization of liquids listed below. |
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Answer» Solution :For ETHANOL : GIVEN : `T_b=78.4^@C=(78.4+273)` =351.4 K `DeltaH_V` (ethanol)=`+42.4 "KJ MOL"^(-1)` `DeltaS_V=(DeltaH_V)/T_b` `DeltaS_V=(+42.4 "kJ mol"^(-1))/"351.4 K"` `DeltaS_V=(+42.400 "J mol"^(-1))/"351.4 K"` `DeltaS_V=+120.66 J K^(-1) "mol"^(-1)` For Toluene : Given : `T_b=110.6^@C`=(110.6+273) =383.6 K `DeltaH_V`(toluene )=+35.2 kJ `"mol"^(-1)` `DeltaS_V=(DeltaH_V)/T_b` `DeltaS_V=(+35.2 "kJ mol"^(-1))/(383.6 K)` `DeltaS_V=(+35200 J mol^(-1))/(383.6 K)` `DeltaS_V=+91.76 J K^(-1) mol^(-1)` |
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