The enthalpy and entropy of vaporization of ammonia at its normal boiling point, -33.34 $\degree$C, are 23.35 kJ/mol and 97.4 J/mol . K, respectively.What is the entropy change of the universe when one mole of liquid ammonia vaporizes in a large room maintained at 34.0 $\degree$C? Assume the final temperature of the ammonia gas is -33.34 $\degree$C.

Question

Quizplus · Accepted Answer

The entropy change of the universe is the sum of the entropy change of the system and the entropy change of the surroundings. The entropy change of the system is given by the entropy of vaporization, which is 97.4 J/mol K. The entropy change of the surroundings is given by the heat capacity of the room multiplied by the change in temperature, which is:$$C_p \Delta T = (20.8 	ext{ J/mol K})(34.0 - 25.0) 	ext{ K} = 187.2 	ext{ J/mol}$$Therefore, the entropy change of the universe is:$$\Delta S_	ext{univ} = \Delta S_	ext{sys} + \Delta S_	ext{surr} = 97.4 	ext{ J/mol K} + 187.2 	ext{ J/mol K} = 284.6 	ext{ J/mol K}$$The correct answer is A.

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

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