Engineering & Science
Trouton’s Rule vs Kistiakowsky Equation: Boiling ΔHvap & ΔSvap
Estimate normal-boiling enthalpy of vaporization ΔH_vap,b (kJ/mol) from normal boiling temperature T_b (K) alone using George Kistiakowsky’s logarithmic entropy law ΔS_vap,b = 36.61 + R * LN(T_b) vs Frederick Trouton’s 88 J/(mol·K) constant.
Why Kistiakowsky’s R·ln(Tb) correction fixes Trouton’s low- and high-boiling drift!
Frederick Trouton (1884) noticed that most non-polar organic liquids vaporize at 1 atm with roughly the same molar entropy jump ΔS_vap,b = ΔH_vap,b / T_b ≈ 88 J/(mol·K) (~10.6 * R) because going from condensed liquid to 1 atm ideal gas expands free volume by roughly the same ratio! Yet for cryogenic gases (like N_2 at 77 K, where 1 atm gas molar volume is tiny) Trouton overpredicts ΔS_vap, while for high-boiling hydrocarbons (500+ K, where 1 atm gas molar volume R*T_b/P is huge) Trouton underpredicts!
George B. Kistiakowsky (1923) incorporated that exact ideal-gas volume expansion R * LN(T_b) into the thermodynamic correlation ΔS_vap,b = 36.61 + 8.3144626 * LN(T_b) J/(mol·K), giving ΔH_vap,b (kJ/mol) = T_b * (36.61 + 8.3144626 * LN(T_b)) / 1000! For benzene (T_b = 353.25 K, experimental ΔH_vap,b = 30.72 kJ/mol): Kistiakowsky gives ΔS_vap = 85.39 J/(mol·K) and ΔH_vap = 30.16 kJ/mol (vs Trouton 31.09 kJ/mol); for n-hexane (T_b = 341.88 K, experimental 28.85 kJ/mol), Kistiakowsky hits 29.10 kJ/mol within 0.9%!
| Non-Polar Liquid & Normal Boiling Point Tb | Kistiakowsky ΔSvap (J/mol·K) | Kistiakowsky ΔHvap (kJ/mol) | Trouton 0.088·Tb Estimate |
|---|---|---|---|
| n-Hexane (Tb = 341.88 K; Expt 28.85) | 85.12 J/(mol·K) | 29.10 kJ/mol (0.9% Err) | 30.09 kJ/mol |
| Benzene (Tb = 353.25 K; Expt 30.72) | 85.39 J/(mol·K) | 30.16 kJ/mol (1.8% Err) | 31.09 kJ/mol |
| Cryogenic Liquid N2 (Tb = 77.36 K; Expt 5.58) | 72.76 J/(mol·K) | 5.63 kJ/mol (0.9% Err!) | 6.81 kJ/mol (+22% Err!) |
Kistiakowsky Predicted Boiling Enthalpy of Vaporization ΔH_vap,b (kJ/mol)
Evaluates T_b * (36.61 + 8.3144626 * LN(T_b)) / 1000 for non-polar and weakly polar fluids (note: strongly hydrogen-bonded alcohols/water exceed both Kistiakowsky and Trouton).
=IF(B2 <= 1, "Invalid Boiling Point Tb", ROUND(B2 * (36.61 + 8.3144626 * LN(B2)) / 1000, 2))Kistiakowsky Molar Entropy of Vaporization ΔS_vap,b (J/(mol·K))
Returns 36.61 + 8.3144626 * LN(T_b) in J/(mol·K) for direct comparison against Trouton’s 88 J/(mol·K) benchmark.
=IF(B2 <= 1, "Invalid Boiling Point Tb", ROUND(36.61 + 8.3144626 * LN(B2), 2))Try the example
Use the original values or change them to test your understanding.