Given: H₂O(l) → H₂O(g) ΔH° = 40.7 kJ at 373K What is the entropy change in the system (ΔS) when one mole of water vaporizes at 100°C and a pressure of one atmosphere?

Question

Quizplus · Accepted Answer

The equation given in the problem represents the enthalpy change when one mole of liquid water (l) vaporizes to form one mole of water vapor (g) at 373K. However, the question asks for the entropy change (ΔS) when one mole of water vaporizes at 100°C and a pressure of one atmosphere.

To find the entropy change, we can use the Clausius-Clapeyron equation:

ΔHvap = TΔSvap

where ΔHvap is the enthalpy of vaporization, T is the boiling point temperature in Kelvin, and ΔSvap is the entropy change of vaporization.

To use this equation, we need to convert the temperature of 100°C to Kelvin:

T = 100°C + 273.15 = 373.15K

We can also use the molar enthalpy of vaporization for water from a thermodynamic table, which is 40.7 kJ/mol, as given in the problem.

Plugging in the values, we get:

40.7 kJ/mol = 373.15K ΔSvap

ΔSvap = 109 J/K/mol.

Note that the entropy change is positive, as expected for a process in which a liquid becomes a gas. This matches with choice C, which is the correct answer.

Given: H2O(l) → H2O(g) ΔH° = 40

Given: H₂O(l) → H₂O(g) ΔH° = 40