Trouton's Rule

Boil a beaker of hexane and a beaker of water. Their boiling points differ enormously, yet the entropy each liquid gains on vaporizing per mole is startlingly similar. This near-constancy is Trouton's Rule, one of the oldest empirical generalizations in physical chemistry. The rule matters because it links a bulk thermodynamic quantity, the enthalpy of vaporization, to something easy to measure, the normal boiling point. It also matters because its failures are diagnostic. Deviations reveal hydrogen bonding, association, or unusual liquid structure. This article derives why Trouton's Rule works, states its numerical value precisely, and shows how to use it to estimate enthalpies of vaporization. You will see the statistical argument behind the constant. Then you will see worked numerical checks. Finally you will see where the rule breaks down and why that breakdown is chemically informative.

Key takeaways

Trouton's Rule: $\Delta_{vap}S \approx 85\ \mathrm{J\,K^{-1}\,mol^{-1}}$ at the normal boiling point, for many "normal" liquids.
It follows because vaporization takes a liquid to a gas of similar molar volume at 1 atm and similar entropy content, for nonpolar, non-associating substances.
Polar or hydrogen-bonded liquids (water, ethanol) show larger $\Delta_{vap}S$ because the liquid is more ordered than a simple liquid.
Very small or very high-boiling substances (helium, metals) deviate the other way, for distinct physical reasons.
The rule lets you estimate $\Delta_{vap}H$ from $T_b$ alone, and estimate $T_b$ from a known $\Delta_{vap}H$ , via the Clausius–Clapeyron framework.
Key idea: at the boiling point, the entropy gained in expanding roughly the same relative volume of gas from a "normal" liquid is nearly universal, about $85\ \mathrm{J\,K^{-1}\,mol^{-1}}$ .

The core insight

Because vaporization at 1 atm converts one mole of liquid into one mole of gas occupying a molar volume that is roughly the same multiple of the liquid's molar volume for most non-associating liquids, the entropy increase per mole at the boiling point is approximately substance-independent, giving the empirical constant $\Delta_{vap}S \approx 85\ \mathrm{J\,K^{-1}\,mol^{-1}}$ .

The physical picture

Vaporization is an order-to-disorder transition: a liquid, with molecules in contact and short-range order, becomes a gas, with molecules essentially independent and randomly distributed through a much larger volume. The entropy of vaporization,

\Delta_{vap}S = \frac{\Delta_{vap}H}{T_b}

measures how much extra disorder is created per unit of thermal energy invested, evaluated at the normal boiling point $T_b$ (1 atm). Frederick Trouton noticed in 1884 that when you compute $\Delta_{vap}S$ for many "ordinary" liquids — nonpolar or weakly polar, non-associating, no hydrogen bonds — the answer clusters tightly around one value. Whether the liquid boils at 111 K (methane) or 353 K (benzene), the entropy gained per mole of vaporization is nearly the same. The reason is structural, not chemical: in each case a comparably disordered liquid becomes a comparably free gas, occupying a similar relative expansion in volume and freedom of motion, at whatever temperature makes vapor pressure equal to 1 atm. The rule is not exact. Substances whose liquid state is more ordered than the simple picture — because of hydrogen bonding (water, alcohols) or strong dipole association — have to shed more entropy-related structure on vaporizing, and so $\Delta_{vap}S$ runs high. Substances with unusually low boiling points and correspondingly weak intermolecular forces (helium, hydrogen, neon) show the opposite deviation, because quantum effects and near-ideal-gas-like liquids change the entropy balance. Very high-boiling liquids, and non-spherical or associated molecules, may also deviate for structural reasons.

Figure

Entropy of vaporization plotted against boiling point for several liquids

Figure

Molecular picture of the liquid-to-gas entropy jump

The core relations

\Delta_{vap}S(T_b) = \frac{\Delta_{vap}H(T_b)}{T_b}

where: $\Delta_{vap}H(T_b)$ is the molar enthalpy of vaporization at the normal boiling point, $T_b$ .

\Delta_{vap}S(T_b) \approx 85\ \mathrm{J\,K^{-1}\,mol^{-1}} \quad \text{(Trouton's Rule)}

where: this value holds for many nonpolar, non-associating liquids; it is empirical, not exact.

\Delta_{vap}H(T_b) \approx (85\ \mathrm{J\,K^{-1}\,mol^{-1}}) \times T_b

where: this rearrangement estimates the heat of vaporization from the boiling point alone.

\ln p = -\frac{\Delta_{vap}H}{RT} + C

where: this is the integrated Clausius–Clapeyron equation, into which the Trouton estimate of $\Delta_{vap}H$ can be substituted to predict vapor pressure curves.

Derivation AWhy the entropy of vaporization is nearly universal

Goal. Show why $\Delta_{vap}S$ at the normal boiling point should be roughly constant across many simple liquids, rather than scattering by orders of magnitude.

Step 1: separate translational and configurational contributions. The molar entropy of a substance can be thought of, loosely, as arising from translational, rotational, and configurational (packing/ordering) freedom. On vaporization at constant $T_b$ , the dominant change is translational: molecules go from being confined to a small liquid volume $V_m^{liq}$ , jostling among neighbors, to occupying a gas volume $V_m^{gas}$ roughly a thousandfold larger, moving nearly independently.

Step 2: appeal to the volume ratio. For an ideal gas, the entropy of expanding one mole from volume $V_1$ to $V_2$ at constant temperature is

\Delta S = R\ln\left(\frac{V_2}{V_1}\right).

At the normal boiling point, essentially every liquid's gas phase occupies about the same molar volume, because all are ideal gases at 1 atm and comparable $T_b$ :

V_m^{gas} = \frac{RT_b}{p} \approx \frac{(8.314)(300)}{101325} \approx 0.0246\ \mathrm{m^3/mol} \approx 25\ \mathrm{L/mol}.

Liquid molar volumes for simple organic and small inorganic liquids are also similar to one another, typically 0.05–0.15 L/mol. So the ratio $V_m^{gas}/V_m^{liq}$ is roughly 200–500 for a wide range of substances — not identical, but confined to a modest range compared with the enormous range of $\Delta_{vap}H$ values themselves.

Step 3: combine with the loss of local order. Beyond the volume term, vaporization also releases the liquid's short-range order (rotational freedom is gained, and any residual configurational order from packing is destroyed). For simple, non-associating liquids this contribution is roughly constant per mole because the local coordination environment in the liquid is similar in kind (van der Waals contacts) even if the interaction strength (and hence $T_b$ ) differs enormously.

Step 4: assemble the constant. Adding a roughly constant translational-expansion term (of order $R\ln(300) \approx 8.314 \times 5.7 \approx 47\ \mathrm{J\,K^{-1}\,mol^{-1}}$ ) to a roughly constant "loss of liquid-like order" term (empirically found to add another ~35–40 J/K/mol) gives a total near

\Delta_{vap}S \approx 85\ \mathrm{J\,K^{-1}\,mol^{-1}}.

Step 5: identify what breaks the constancy. The derivation assumed the liquid is a simple, unstructured, van der Waals liquid. If instead the liquid has strong directional bonding (hydrogen bonds in water or alcohols), its entropy is already anomalously low relative to a simple liquid of the same $T_b$ . Vaporization must then supply a larger entropy increase to reach the same disordered-gas reference state, so $\Delta_{vap}S$ exceeds 85 J/K/mol. Conversely, if intermolecular forces are so weak that even the liquid is nearly gas-like in its disorder (helium, hydrogen), less entropy needs to be gained, so $\Delta_{vap}S$ falls below 85 J/K/mol.

\boxed{\Delta_{vap}S(T_b) \approx 85\ \mathrm{J\,K^{-1}\,mol^{-1}} \text{ for simple, non-associating liquids}}

Worked examples

Example 1 — Foundational: checking Trouton's Rule for benzene. Benzene has a normal boiling point $T_b = 353.25\ \mathrm{K}$ and a measured molar enthalpy of vaporization $\Delta_{vap}H = 30.8\ \mathrm{kJ/mol}$ at that temperature. Step 1. Compute $\Delta_{vap}S$ directly:

\Delta_{vap}S = \frac{\Delta_{vap}H}{T_b} = \frac{30{,}800\ \mathrm{J/mol}}{353.25\ \mathrm{K}} = 87.2\ \mathrm{J\,K^{-1}\,mol^{-1}}.

Step 2. Compare with the Trouton value of 85 J/K/mol:

\text{deviation} = \frac{87.2 - 85}{85} \times 100\% \approx 2.6\%.

This is well within the scatter typical of the rule (roughly ±10–15% for "normal" liquids), confirming benzene behaves as expected for a nonpolar, non-associating liquid. Sanity check: benzene has no hydrogen bonds and modest dipole moment (0.0 D by symmetry, though it has polarizable π electrons), consistent with near-ideal Trouton behavior.

Example 2 — Applied: predicting a boiling point and testing water's anomaly. Suppose you are given that a newly synthesized nonpolar liquid has $\Delta_{vap}H = 28.0\ \mathrm{kJ/mol}$ , and you want to estimate its normal boiling point using Trouton's Rule, then check the prediction's plausibility against water, a known anomalous case. Step 1. Rearrange Trouton's Rule to estimate $T_b$ :

T_b \approx \frac{\Delta_{vap}H}{85\ \mathrm{J\,K^{-1}\,mol^{-1}}} = \frac{28{,}000\ \mathrm{J/mol}}{85\ \mathrm{J\,K^{-1}\,mol^{-1}}} = 329\ \mathrm{K}.

This is a physically reasonable boiling point (about 56 °C), consistent with a small-to-medium organic liquid such as a ketone or chlorinated solvent. Step 2. Now apply the same procedure incorrectly to water, to see the rule fail, and quantify the failure. Water's true values are $T_b = 373.15\ \mathrm{K}$ and $\Delta_{vap}H = 40.7\ \mathrm{kJ/mol}$ . Naively estimating $\Delta_{vap}H$ from Trouton's Rule using the true $T_b$ :

\Delta_{vap}H_{\text{Trouton}} \approx (85\ \mathrm{J\,K^{-1}\,mol^{-1}})(373.15\ \mathrm{K}) = 31{,}700\ \mathrm{J/mol} = 31.7\ \mathrm{kJ/mol}.

Step 3. Compare with the true value:

\text{error} = \frac{40.7 - 31.7}{40.7}\times 100\% \approx 22\%,

a substantial underestimate. The true $\Delta_{vap}S$ for water is

\Delta_{vap}S = \frac{40{,}700}{373.15} = 109.1\ \mathrm{J\,K^{-1}\,mol^{-1}},

about 28% above the Trouton value. This is exactly the expected direction and rough magnitude of deviation for a strongly hydrogen-bonded liquid: water's liquid state is far more ordered (each molecule hydrogen-bonds to ~3.5 neighbors on average) than a simple van der Waals liquid, so vaporization must supply extra entropy to reach the disordered gas state. The sanity check here is qualitative but essential: knowing water hydrogen-bonds tells you in advance which direction the rule will fail, before doing any arithmetic.

Limiting cases & common traps

Situation	Behavior	Why
Nonpolar, non-associating liquids (Ar, CCl4, C6H6, hydrocarbons)	$\Delta_{vap}S \approx 80$ – $90\ \mathrm{J/K/mol}$	Simple van der Waals liquid structure, as assumed in the derivation
Hydrogen-bonded liquids (H2O, alcohols, carboxylic acids)	$\Delta_{vap}S$ well above 85, often 100–120	Liquid is more ordered than "simple"; extra entropy must be supplied
Very low-boiling, weakly interacting gases (He, H2, Ne)	$\Delta_{vap}S$ well below 85	Liquid is already close to a disordered, quantum, or nearly ideal state
Associating liquids that dimerize in the vapor (acetic acid)	$\Delta_{vap}S$ below 85	Gas phase is less disordered than expected because dimers persist
High-boiling metals and ionic liquids	Often deviate, sometimes obey a modified rule	Different bonding type (metallic/ionic) changes both liquid and vapor entropy content

The chemist behind it

Frederick Thomas Trouton reported the regularity in 1884 while a young physicist working under George FitzGerald in Dublin, tabulating latent heats and boiling points for dozens of liquids. He noticed that dividing molar latent heat of vaporization by boiling point gave a nearly constant number across chemically dissimilar substances — a striking early example of a dimensional, structure-based generalization emerging from tabulated data long before statistical mechanics could explain it. The explanation in terms of comparable gas-phase molar volumes and comparable loss of liquid order came only decades later, once entropy and molecular statistics were better understood.

Origins · 1 minFrederick Thomas Trouton and how this was discovered

Frederick Thomas Trouton

1863–1922 · Irish physicist who first tabulated the empirical constancy of ΔvapS/Tb across liquids in 1884

Ask the tutor: How does the Clausius–Clapeyron equation combine with Trouton's Rule to estimate an entire vapor-pressure curve from just one boiling point?

Why does Trouton's Rule use the normal boiling point specifically, rather than any temperature

The normal boiling point is the temperature at which vapor pressure equals exactly 1 atm, so the gas-phase molar volume is fixed at $RT_b/p$ for every substance being compared, at its own $T_b$ . This standardization is what makes the volume-ratio argument in the derivation work: all liquids are compared under matched physical conditions, gas at 1 atm. Using an arbitrary temperature would compare gases at different pressures and volumes, destroying the near-constancy the rule relies on.

Can Trouton's Rule be used to estimate boiling points of unknown compounds

Yes, in a rough, order-of-magnitude sense, provided the compound is nonpolar and non-associating. Given an estimated or measured $\Delta_{vap}H$ , dividing by 85 J/K/mol gives a plausible $T_b$ , useful for sanity-checking distillation conditions or comparing homologous series. It should not be trusted to better than about ±10–20%, and it fails outright for hydrogen-bonded or ionic substances, as demonstrated for water.

Is there a modified or improved version of Trouton's Rule

Yes. Hildebrand's Rule and related corrections normalize for the fact that boiling points differ partly because gas-phase entropies differ slightly with molar volume at the boiling temperature itself. One refinement compares $\Delta_{vap}S$ at the temperature where each vapor occupies the same molar volume (rather than each at its own 1 atm boiling point), improving agreement for a wider range of nonpolar liquids, though hydrogen-bonded liquids still deviate.

Does Trouton's Rule apply to sublimation or to solid-solid transitions

No. Trouton's Rule specifically concerns the liquid-to-gas transition at the normal boiling point, where the entropy change is dominated by the enormous, fairly universal expansion in volume and loss of short-range liquid order. Sublimation (solid to gas) involves a different, generally larger entropy change because the solid is more ordered than the liquid, and solid-solid transitions have no comparable universal volume argument at all.

Why do carboxylic acids like acetic acid deviate below the Trouton value

Carboxylic acids partially dimerize through double hydrogen bonds even in the vapor phase, particularly near the boiling point. Because the gas is not a fully disordered, monomeric ideal gas, its entropy is lower than the simple picture assumes, so the overall entropy gained on vaporizing is smaller than 85 J/K/mol. This is a case where structure persists on both sides of the phase transition, changing the balance in the opposite direction from water's case.