SME
General Chemistry II · Problem type
Entropy of vaporization
Use this for “Ethanol boils at 351.5 K, and its heat of vaporization is 38.6 kJ/mol. Find its entropy of vaporization.”
Chart
ΔGᵥₐₚ(352) = 0
This is an example. Type your own values to start a new problem.
ΔSᵥₐₚ (J/(mol·K))
=
ΔHᵥₐₚ (kJ/mol)
÷
Tb (K)
Formulas
ΔSᵥₐₚ
=
1,000
×
ΔHᵥₐₚ
Tb
110
=
1,000
×
38.6
351.5
Assumptions
•
At the normal boiling point the liquid and its vapor are at equilibrium, so ΔGᵥₐₚ = ΔHᵥₐₚ − TbΔS_vap = 0 and ΔSᵥₐₚ = ΔHᵥₐₚ ÷ Tb.
•
ΔHᵥₐₚ is in kJ/mol and ΔSᵥₐₚ in J/(mol·K), so ΔHᵥₐₚ is multiplied by 1000. Tb is in kelvins (°C + 273.15).
•
Trouton’s rule: most liquids give about 85–88 J/(mol·K). Hydrogen-bonded liquids such as water give more, because their liquid is more ordered.
•
Melting works the same way: ΔSfus = ΔHfus ÷ Tₘ at the melting point.
Step-by-step
WE KNOW
ΔHᵥₐₚ = 38.6 kJ/mol, Tb = 351.5 K
FIND
entropy of vaporization (ΔSᵥₐₚ)
Step 1 · Find entropy of vaporization (ΔSᵥₐₚ)
Use ΔSᵥₐₚ = 1,000 × ΔHᵥₐₚ ÷ Tb
At the boiling point ΔG = 0, so ΔH = TbΔS: divide the heat, in J/mol, by the boiling point.
ΔSᵥₐₚ
=
1,000
×
38.6
351.5
ΔSᵥₐₚ
=
38,600
351.5
ΔSᵥₐₚ = 110 J/(mol·K)
Check
Put the numbers back in. Both sides should match.
110
=
1,000
×
38.6
351.5
✓
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Entropy of the universe (second law)
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