Question 10.4: CALCULATING AN ENTROPY OF VAPORIZATION The boiling point of ...
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 ΔH_{vap} = 40.67 kJ/mol. What is the entropy change for vaporization, ΔS_{vap}, 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.
ΔS_{vap} = \frac{ΔH_{vap}}{T} = \frac{40.67\frac{kJ}{mol}}{373.15 K} = 0.1090 kJ/(K · mol) = 109.0 J/(K · mol)
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.