The molar enthalpychange of vaporizing ethanol is 38.6 kJ/mol at the boilingpoin

Question

The molar enthalpychange of vaporizing ethanol is 38.6 kJ/mol at the boilingpoint

(78.4 °C);likewise, it takes 23.35 kJ/mol for ammonia to boil at33.34 °C. If you

calculateS for one mole of each of these substances,would you say that the values are

very different from each other? Why might thatbe?

Hint: The entropy ofa liquid is much, much smaller than a gas, and you arebasically

calculatingSgas-Sliq. Also, do perfect gases have very differententropies if P, T and V are

the same (or verysimilar)?

Explanation / Answer

We know that vapS = vapH /Tb For ethanol, vapS = 38.6kJ/mol /(78.4+273)K   = 109.85J/K For ammonia, vapS = 23.35kJ/mol /(-33.34+273)K = 97.43J/K Ethanol associated with hydrogen bonding, which requires moreamount of energy to overcome these forces of attraction. For ethanol, vapS = 38.6kJ/mol /(78.4+273)K   = 109.85J/K For ammonia, vapS = 23.35kJ/mol /(-33.34+273)K = 97.43J/K Ethanol associated with hydrogen bonding, which requires moreamount of energy to overcome these forces of attraction.

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