Calculating an Entropy of Vaporization The boiling point of water is 100 °C, and the enthalpy change for the conversion of water to steam is ΔHvap = 40.67 kJ/mol. What is the entropy change for vaporization, ΔSvap, in J/(K · mol)?

Question

Calculating an Entropy of Vaporization

The boiling point of water is 100 °C, and the enthalpy change for the conversion of water to steam is Δ[latex]H_{vap}[/latex] = 40.67 kJ/mol. What is the entropy change for vaporization, Δ[latex]S_{vap}[/latex], in J/(K · mol)?

STRATEGY

At the temperature where a phase change occurs, the two phases coexist in equilibrium and ΔG, the free-energy difference between the phases, is zero: ΔG = ΔH - TΔS = 0. Rearranging this equation gives ΔS = ΔH/T, where both ΔH and T are known. Remember that T must be expressed in kelvin.

Accepted Answer

[latex]\Delta S_{ ext {vap }}=\frac{\Delta H_{ ext {vap }}}{T}=\frac{40.67 \frac{\mathrm{kJ}}{\mathrm{mol}}}{373.15 \mathrm{~K}}=0.1090 \mathrm{~kJ} /(\mathrm{K} \cdot \mathrm{mol})=109.0 \mathrm{~J} /(\mathrm{K} \cdot \mathrm{mol})[/latex]

As you might expect, there is a large positive entropy change, corresponding to a large increase in randomness, on converting water from a liquid to a gas.

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