Trouton's Rule: Entropy of Vaporisation and Estimating ΔHvap

By Aniket Bhardwaj · 24 October 2026 · Physical Chemistry

Trouton’s rule is a short empirical statement with a deep reason behind it: for many ordinary liquids, the entropy of vaporisation at the normal boiling point is almost the same number. In IIT-JAM, GATE and CSIR-NET physical chemistry it appears in three ways: as a one-line estimate of ΔH_vap from a boiling point, as a test question on why water and ethanol break the rule, and as the first step in a Clausius–Clapeyron numerical where ΔH_vap is not given.

The rule

ΔS_vap = ΔH_vap ÷ T_b ≈ 85–88 J K⁻¹ mol⁻¹
Estimate: ΔH_vap ≈ (88 J K⁻¹ mol⁻¹) × T_b

T_b is the normal boiling point in kelvin (the boiling point at 1 atm, or at the standard pressure your book specifies). ΔH_vap is the molar enthalpy of vaporisation in J mol⁻¹. Be careful with units: ΔH_vap is usually printed in kJ mol⁻¹, and ΔS_vap comes out in J K⁻¹ mol⁻¹ only after you multiply by 1000.

The constant is quoted as about 85 in some books and about 88 in others; some write 10.5R, which is about 87.3 J K⁻¹ mol⁻¹. Use the value in your syllabus book or the value given in the question.

Where it comes from

At the boiling point, liquid and vapour are in equilibrium, so ΔG = 0 and ΔS_vap = ΔH_vap/T_b exactly. That part is thermodynamics. The surprise is that the number is similar for many liquids. The reason is that the vapour of any liquid, compared at the normal boiling point, occupies a large and similar volume (at 1 atm and 353 K an ideal gas occupies RT/P = 29 litres per mole), while the liquid occupies only a small volume (tens of mL per mole). Going from a small, ordered volume to a large, disordered one produces a similar gain in disorder for simple liquids. Compare liquids at the same pressure and you compare similar vapour volumes.

Worked example 1: testing the rule on benzene

Data (typical textbook values): benzene, T_b = 353.2 K, ΔH_vap = 30.8 kJ mol⁻¹.

ΔS_vap = 30 800 J mol⁻¹ ÷ 353.2 K = 87.2 J K⁻¹ mol⁻¹.

This sits inside the 85–88 range, so benzene obeys Trouton’s rule well. The liquid is non-polar and has no strong specific attractions, which is the case the rule describes best.

Worked example 2: liquids that do not obey it

Water: T_b = 373.15 K, ΔH_vap = 40.65 kJ mol⁻¹. ΔS_vap = 40 650 ÷ 373.15 = 108.9 J K⁻¹ mol⁻¹.

Ethanol: T_b = 351.5 K, ΔH_vap = 38.6 kJ mol⁻¹. ΔS_vap = 38 600 ÷ 351.5 = 109.8 J K⁻¹ mol⁻¹.

Mercury: T_b = 629.9 K, ΔH_vap = 59.1 kJ mol⁻¹. ΔS_vap = 59 100 ÷ 629.9 = 93.8 J K⁻¹ mol⁻¹.

Water and ethanol are about 20 J K⁻¹ mol⁻¹ above the rule. Their liquid state is more ordered than a simple liquid because of hydrogen bonding, so more entropy is gained when the order is destroyed on vaporisation. Mercury is above the rule too; a liquid metal has a different kind of structure from a molecular liquid, and the simple argument about similar disorder does not hold. In general the rule fails for strongly associated liquids.

Worked example 3: estimating ΔH_vap and then a boiling point at lower pressure

Question. A non-polar liquid boils at 353.2 K at 1 atm. Its ΔH_vap is not given. Estimate ΔH_vap, and find its boiling point when the pressure is reduced to 0.50 atm. Take R = 8.314 J K⁻¹ mol⁻¹ and use Trouton’s constant 88 J K⁻¹ mol⁻¹.

Step 1. ΔH_vap ≈ 88 × 353.2 = 31 082 J mol⁻¹ (about 31.1 kJ mol⁻¹).

Step 2. Clausius–Clapeyron: ln(P₂/P₁) = −(ΔH_vap/R)(1/T₂ − 1/T₁). With P₂/P₁ = 0.50, ln 0.50 = −0.6931.

Step 3. Rearrange: 1/T₂ = 1/T₁ − R ln(P₂/P₁)/ΔH_vap = 1/353.2 + (8.314 × 0.6931)/31 082.

1/353.2 = 0.0028313. The second term = 5.7626 ÷ 31 082 = 0.00018540. Sum = 0.0030167.

Step 4. T₂ = 1/0.0030167 = 331.5 K, which is 58.3 °C.

Check: with the real ΔH_vap of 30.8 kJ mol⁻¹ the answer is 331.3 K. The Trouton estimate is within about 0.2 K here, so the shortcut is good enough for a liquid that obeys the rule. For water the same shortcut would underestimate ΔH_vap by roughly 20%, so never use it there.

Worked example 4: how large is the error for water?

Trouton estimate for water: 88 × 373.15 = 32 837 J mol⁻¹ = 32.8 kJ mol⁻¹. Experimental value used above: 40.65 kJ mol⁻¹.

Difference = 40.65 − 32.84 = 7.81 kJ mol⁻¹, which is 7.81 ÷ 40.65 = 19% of the real value.

Common mistakes

Using °C for T_b. The rule needs kelvin. 80 °C is 353 K, not 80.
Mixing kJ and J. 30.8 kJ mol⁻¹ must become 30 800 J mol⁻¹ before dividing to get J K⁻¹ mol⁻¹.
Applying it to hydrogen-bonded liquids. Water, alcohols and carboxylic acids show larger values; the question usually expects you to say so.
Using a boiling point at a non-standard pressure. The rule is for the normal boiling point.
Treating it as a law. It is an empirical rule with a thermodynamic explanation, not an exact relation. The exact relation is ΔS = ΔH/T at equilibrium.
Sign errors in Clausius–Clapeyron when you combine the two ideas; check that lower pressure gives a lower boiling point.

Exam relevance

Question type	What to do
Estimate ΔH_vap from T_b	ΔH_vap ≈ constant × T_b, T in kelvin
Why does water deviate?	Hydrogen bonding makes the liquid more ordered, so ΔS_vap is larger
Boiling point at reduced pressure with ΔH_vap missing	Trouton estimate, then Clausius–Clapeyron
Assertion–reason on entropy of vaporisation	ΔS_vap = ΔH_vap/T_b because ΔG = 0 at equilibrium
Comparing liquids	Larger ΔS_vap than 88 suggests association in the liquid

Exam patterns and marks change from year to year. Check the current official syllabus and notification for your exam rather than relying on any list in an article.

Use the Clausius–Clapeyron tool to repeat Example 3: enter the two pressures, one temperature and the estimated ΔH_vap, and check the second temperature.

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