The enthalpy of vaporization of water at 100.0 $\circ$ C is 40.68 kJ/mol.What is the entropy change in the surroundings when one mole of water vapor condenses at 100.0 $\circ$ C in a large room maintained at a temperature of 25.00 $\circ$ C?

Question

Quizplus · Accepted Answer

The entropy change in the surroundings can be calculated using the equation:

ΔS = -ΔH/T

where ΔH is the enthalpy change (the enthalpy of vaporization, 40.68 kJ/mol), and T is the temperature of the surroundings (25.00°C = 298.15 K).

Substituting the values:

ΔS = -40.68 kJ/mol / 298.15 K 
ΔS = -136.4 J/K

Since the entropy change is negative, the surroundings become more ordered (i.e. less disordered) as heat is released during the condensation of water vapor. The correct answer is thus (E) +136.4 J/K, which reflects the magnitude of the entropy change.

The Enthalpy of Vaporization of Water at 100 ∘\circ∘ C Is 40

The Enthalpy of Vaporization of Water at 100 $\circ$ C Is 40