The entropy of vaporization of water is 109.0 J/mol · K. What is the enthalpy of vaporization of water at its normal boiling point?

Question

Quizplus · Accepted Answer

The Clausius-Clapeyron equation relates the enthalpy of vaporization, the melting point, and the boiling point of a substance. It is given by:

ln(P2/P1) = ∆Hvap/R [(1/T1) - (1/T2)]

where P1 and P2 are the vapor pressures at temperatures T1 and T2 respectively, R is the gas constant, and ∆Hvap is the enthalpy of vaporization.

At the normal boiling point of water (100°C or 373 K), the vapor pressure is 1 atm. At the melting point of water (0°C or 273 K), the vapor pressure is negligible.

Therefore, we can rearrange the Clausius-Clapeyron equation as:

∆Hvap = - R [(ln(P2/P1))/(1/T2 - 1/T1)]

Substituting the given values, we get:

∆Hvap = - (8.314 J/mol·K) [(ln(1 atm/0))/(1/373 K - 1/273 K)] = -40.66 kJ/mol

The negative sign indicates that the process of vaporization is exothermic (i.e. releases heat), which is expected as water molecules come together in the liquid state and break apart in the vapor state. Therefore, the best choice is D.

The Entropy of Vaporization of Water Is 109