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 enthalpy of vaporization (ΔHvap) can be calculated using the Clausius-Clapeyron equation:

ln(P2/P1) = (ΔHvap/R)(1/T1 - 1/T2)

At the normal boiling point of water, the temperature (T1) is 373 K and the vapor pressure (P1) is 1 atm. At any temperature below the boiling point, the vapor pressure (P2) can be assumed to be negligible or zero. So, we can simplify the equation as:

ln(0) = (ΔHvap/R)(1/373 - 0)

The natural logarithm of zero is undefined, indicating that the equation cannot be solved for ΔHvap. However, we can use the given entropy of vaporization (ΔSvap) to estimate ΔHvap. The relationship between ΔHvap and ΔSvap is:

ΔHvap = TΔSvap

Substituting the values:

ΔHvap = (373 K)(109.0 J/mol · K) = 40.66 kJ/mol

Therefore, the answer is D, +40.66 kJ/mol.

The Entropy of Vaporization of Water Is 109