Question 11.2: 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 ∆H_{vap} = 40.67 kJ/mol. What is the entropy change for vaporization, ∆S_{vap} , in J/(K . mol)?
IDENTIFY
| Known | Unknown |
| Boiling point of water (T = 100 °C) | Entropy change (∆S_{vap}) |
| Enthalpy change (∆H_{vap} = 40.67 kJ/mol) | |
STRATEGY
At the temperature where a phase change occurs, the two phases coexist in equilibrium and ∆G, the free-energy difference between the phases, is zero: ∆G = ∆H – T∆S = 0. Rearranging gives ∆S = ∆H/T, where both ∆H and T are known. Remember that T must be expressed in kelvin.
∆S_{vap}=\frac{∆H_{vap}}{T}=\frac{40.67\frac{kJ}{mol} }{373.15 K}=0.1090 kJ/(K.mol)=109.0 J/(K.mol)
CHECK
Converting water from liquid to a gas should result in a large and positive entropy change. The calculation for ∆S_{vap} agrees with the expected sign and magnitude and is therefore a reasonable answer.