The enthalpy and entropy of vaporization of ammonia at its normal boiling point, -33.34$\degree$C, are approximately 23.35 kJ/mol and 97.4 J/mol . K, respectively.What would the entropy change of the universe be if one mole of ammonia vapor were to condense in a large room maintained at 30.0$\degree$C?

Question

Quizplus · Accepted Answer

The entropy change of the universe is the sum of the entropy change of the system and the surroundings. For condensation, the system's entropy change is negative, equal to the negative of the entropy of vaporization (-97.4 J/K). The surroundings' entropy change is calculated using ΔS = q/T, where q is the heat released (23.35 kJ/mol) and T is the temperature in Kelvin (303.15 K). The total entropy change of the universe is negative, approximately -20.4 J/K.

The Enthalpy and Entropy of Vaporization of Ammonia at Its °\degree°

The Enthalpy and Entropy of Vaporization of Ammonia at Its $\degree$