Entropy of Phase Transitions

Ice melts at exactly 0 °C under one atmosphere, absorbing heat without changing temperature. That stillness hides a jump in molecular disorder that we can measure precisely. Because melting and boiling happen at constant temperature and (usually) constant pressure, the entropy change is simply the heat absorbed divided by that fixed temperature. This idea matters because phase transitions are the cleanest laboratory for the thermodynamic definition of entropy. There is no need to integrate heat capacities over a changing temperature. You just need one enthalpy and one temperature, both measurable by calorimetry. This article derives the entropy of fusion and vaporization from transition enthalpies. It works two full numerical examples. It also previews Trouton's Rule and the calorimetric route to absolute entropies, which build directly on these results.

Key takeaways

At a phase transition, $T$ is constant, so $\Delta S = \Delta H_{trs}/T_{trs}$ exactly.
This works because phase transitions at their normal transition temperature are reversible, constant-pressure processes.
Entropy of vaporization is almost always much larger than entropy of fusion for the same substance.
Trouton's Rule says $\Delta_{vap}S \approx 85$ J/K per mole for many "normal" liquids at their boiling point.
Hydrogen-bonded or associated liquids (water, ethanol) violate Trouton's Rule because their vapors are less disordered relative to the liquid.
Key idea: divide the measured transition enthalpy by the transition temperature to get the entropy change, because $q_{rev} = \Delta H$ at constant $P$ and $T$ is fixed during the transition.

The core insight

Because a phase transition at its equilibrium temperature is a reversible, isothermal, constant-pressure process, the heat exchanged equals the enthalpy of transition exactly, so the entropy change collapses to the ratio $\Delta H_{trs}$ over $T_{trs}$ , with no integral required.

The physical picture

Picture a solid–liquid mixture sitting at exactly the melting point, with heat trickling in from a surroundings held at that same temperature. Because both phases coexist, adding heat does not raise the temperature. It converts solid into liquid instead. This is the defining feature of a first-order phase transition: latent heat flows at constant $T$ . Recall that entropy is defined thermodynamically by

dS = \frac{dq_{rev}}{T}

For any reversible process. A phase change carried out infinitesimally slowly, with the two phases in equilibrium at temperature $T_{trs}$ and pressure held at 1 bar (the standard state), is reversible in exactly this sense. The system moves along an equilibrium line, never departing from a state definable by $(P,T)$ alone. Since $T$ does not change during the transition, it can be pulled out of the integral:

\Delta S_{trs} = \int \frac{dq_{rev}}{T} = \frac{1}{T_{trs}}\int dq_{rev} = \frac{q_{rev}}{T_{trs}}

At constant pressure, $q_{rev} = \Delta H_{trs}$ , the enthalpy of transition — the same quantity you would measure by simple calorimetry (heat supplied per mole, at constant pressure, while temperature holds steady). No heat capacity integration is needed, unlike the entropy change on heating a single phase, where $C_P$ must be integrated over a changing $T$ . Physically, melting and vaporization both increase the number of accessible microstates: molecules gain translational and orientational freedom. Vaporization typically produces a far bigger jump, because a gas occupies roughly a thousand times the volume of the condensed phase and molecules become essentially independent. Fusion, by contrast, retains short-range order and strong intermolecular contacts; the solid-to-liquid step is a much gentler loosening.

Figure

Entropy change along a constant-pressure heating and phase-change path

Figure

Typical magnitudes for water: fusion versus vaporization

The core relations

\Delta_{fus}S = \frac{\Delta_{fus}H}{T_f}

where $\Delta_{fus}H$ is the molar enthalpy of fusion (always positive) and $T_f$ is the normal melting point in kelvin.

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

where $\Delta_{vap}H$ is the molar enthalpy of vaporization and $T_b$ is the normal boiling point in kelvin.

\Delta_{trs}S = \frac{\Delta_{trs}H}{T_{trs}}, \qquad \Delta_{fus}S(T_b \to T_f\ \text{path}) = -\Delta_{freeze}S

where the second relation records that freezing at $T_f$ is simply the reverse process, so $\Delta_{freeze}S = -\Delta_{fus}S$ .

\Delta S_{surr} = -\frac{\Delta H_{trs}}{T_{trs}}, \qquad \Delta S_{total} = \Delta S_{trs} + \Delta S_{surr} = 0

where $\Delta S_{surr}$ is the surroundings' entropy change; the total is zero because the transition at $T_{trs}$ is reversible (system and surroundings at equilibrium).

Derivation AFrom the Clausius definition to ΔS = ΔH/T

Goal. Show rigorously that the entropy change accompanying a phase transition at its equilibrium temperature equals the transition enthalpy divided by that temperature.

Step 1. Set up the process. Consider one mole of a pure substance converting from phase α to phase β (e.g., solid to liquid) at constant pressure $P^{\ominus}$ , held at the equilibrium transition temperature $T_{trs}(P^{\ominus})$ — the unique temperature at which α and β have equal chemical potential at that pressure.

Step 2. Confirm reversibility. At $T_{trs}$ , the two phases are in mutual equilibrium: an infinitesimal quantity of heat can convert a little α into β, or be removed to convert β back into α, with no net driving force beyond the infinitesimal push applied. This is, by definition, a reversible process. It is this equilibrium condition — not merely "constant temperature" — that licenses use of $dS = dq_{rev}/T$ .

Step 3. Apply the Clausius definition. The entropy change of the system is

\Delta S = \int_{\alpha}^{\beta} \frac{dq_{rev}}{T}

Step 4. Pull out the constant temperature. Because the transition proceeds entirely at $T = T_{trs}$ , this constant factors outside the integral:

\Delta S = \frac{1}{T_{trs}} \int_{\alpha}^{\beta} dq_{rev} = \frac{q_{rev}}{T_{trs}}

Step 5. Identify $q_{rev}$ with the enthalpy of transition. At constant pressure with only $PV$ work, the first law gives $dU = dq - PdV$ , and enthalpy is defined as $H = U + PV$ , so $dH = dq_{rev}$ at constant $P$ . Integrating over the whole transition,

q_{rev} = \Delta H_{trs}

Step 6. Combine. Substituting Step 5 into Step 4 gives the target relation. The same logic applies unchanged to vaporization, sublimation, or any solid–solid polymorphic transition, provided $T_{trs}$ is the genuine equilibrium temperature at the stated pressure.

\boxed{\Delta S_{trs} = \frac{\Delta H_{trs}}{T_{trs}}}

Worked examples

Example 1 — Melting and boiling of water (foundational). Data: $\Delta_{fus}H^{\ominus} = 6.01$ kJ/mol at $T_f = 273.15$ K; $\Delta_{vap}H^{\ominus} = 40.7$ kJ/mol at $T_b = 373.15$ K (both at 1 atm, standard reference values). Entropy of fusion:

\Delta_{fus}S = \frac{6010\ \text{J/mol}}{273.15\ \text{K}} = 22.0\ \text{J K}^{-1}\text{mol}^{-1}

Entropy of vaporization:

\Delta_{vap}S = \frac{40700\ \text{J/mol}}{373.15\ \text{K}} = 109.1\ \text{J K}^{-1}\text{mol}^{-1}

Sanity check: both are positive, consistent with disorder increasing on melting and on boiling. The vaporization entropy is roughly five times the fusion entropy — expected, since gas formation liberates far more positional and orientational freedom than melting does. Note water's $\Delta_{vap}S$ (109 J/K·mol) sits noticeably above the ~85 J/K·mol "normal liquid" value from Trouton's Rule (previewed below), a signature of hydrogen bonding that leaves the liquid more ordered than typical van der Waals liquids. We can also check consistency with reverse-process reasoning: freezing water at 273.15 K releases 6.01 kJ/mol, so $\Delta_{freeze}S = -22.0$ J/K·mol, and the surroundings (a large reservoir at 273.15 K) gain $+22.0$ J/K·mol — the total entropy change for freezing at the equilibrium point is exactly zero, confirming reversibility.

Example 2 — Benzene vaporization and using entropy of vaporization to back out an unknown enthalpy (applied). Benzene boils at $T_b = 353.25$ K (80.1 °C) with $\Delta_{vap}H^{\ominus} = 30.8$ kJ/mol. Step A: compute $\Delta_{vap}S$ directly.

\Delta_{vap}S = \frac{30800\ \text{J/mol}}{353.25\ \text{K}} = 87.2\ \text{J K}^{-1}\text{mol}^{-1}

This value sits close to the Trouton constant of ≈85 J/K·mol, consistent with benzene being a "normal," non-associating liquid with no hydrogen bonding — a good candidate for Trouton's Rule. Step B: suppose instead we are given only that a similar nonpolar liquid, call it X, boils at $T_b = 300.0$ K and is expected (by Trouton's Rule) to have $\Delta_{vap}S \approx 85$ J/K·mol. Estimate its enthalpy of vaporization.

\Delta_{vap}H \approx T_b \times \Delta_{vap}S = 300.0\ \text{K} \times 85\ \text{J K}^{-1}\text{mol}^{-1} = 25\,500\ \text{J/mol} \approx 25.5\ \text{kJ/mol}

Sanity check: this falls in the typical 20–35 kJ/mol range for small nonpolar organic liquids (compare benzene's 30.8 kJ/mol and toluene's 33.2 kJ/mol), so the estimate is physically reasonable. This inverse use — predicting $\Delta H$ from an assumed "normal" $\Delta S$ — is exactly the logic behind Trouton's Rule as a practical estimation tool, covered next.

Limiting cases & common traps

Situation	What to watch for
Using $\Delta S = \Delta H/T$ away from $T_{trs}$	Invalid — the phases are not in equilibrium off the transition temperature; the process is no longer reversible at that $T$ .
Sublimation	Same formula applies: $\Delta_{sub}S = \Delta_{sub}H / T_{sub}$ , with $\Delta_{sub}H \approx \Delta_{fus}H + \Delta_{vap}H$ near the triple point.
Non-standard pressure	$T_{trs}$ shifts with pressure (Clausius–Clapeyron); always match $\Delta H$ and $T$ at the same pressure.
Comparing fusion vs. vaporization entropies	Fusion entropies vary widely (5–40 J/K·mol) depending on how ordered the solid was; vaporization entropies cluster more tightly (Trouton's Rule) because all normal gases reach comparably high disorder.
Associated liquids (water, alcohols, carboxylic acids)	$\Delta_{vap}S$ exceeds ~85 J/K·mol because hydrogen bonding suppresses disorder in the liquid state, making the liquid-to-gas jump unusually large.

The chemist behind it

Rudolf Clausius introduced both the entropy concept and its defining relation $dS = dq_{rev}/T$ in the 1850s and 1860s while formalizing the Second Law from Carnot's work on heat engines. His formulation gave chemists the tool to convert calorimetric heat measurements — enthalpies of fusion and vaporization, measured directly with calorimeters — into a state function governing spontaneity and equilibrium. The application to ordinary phase changes, so simple in hindsight, was a direct and immediate consequence of his general definition, applied to the special reversible case of two phases in equilibrium.

Origins · 1 minRudolf Clausius and how this was discovered

Rudolf Clausius

1822–1888 · Formulated entropy and the mathematical statement of the Second Law of Thermodynamics

Ask the tutor: How do we handle entropy calculations when a substance passes through a phase transition while being heated from near 0 K to well above its boiling point?

Why is the entropy of fusion usually much smaller than the entropy of vaporization?

Melting only breaks the long-range order of a crystal lattice while keeping molecules close together with similar neighbors and interaction energies; the liquid retains substantial short-range structure. Vaporization instead separates molecules to distances where interactions become negligible, and the resulting gas explores enormously more translational and rotational microstates. Quantitatively, $\Delta_{vap}S$ is typically 3 to 10 times larger than $\Delta_{fus}S$ for the same substance, as seen in water (109 vs. 22 J/K·mol) and most simple molecular liquids.

Does the formula ΔS = ΔH/T require the process to be carried out reversibly in the lab?

No — entropy is a state function, so $\Delta S$ depends only on initial and final states, not on how the transition is actually run. The formula uses the reversible heat only as a calculation device: since melting a solid at $T_f$ under equilibrium conditions is a reversible path between the same two states, its $q_{rev}/T$ gives the correct $\Delta S$ regardless of whether your real sample melted quickly, unevenly, or with a slight superheating in practice.

Can this formula be used for solid-to-solid phase transitions, like graphite to diamond?

Yes, provided $T_{trs}$ is the genuine equilibrium temperature between the two solid phases at the stated pressure, and $\Delta H_{trs}$ is measured at that same condition. Solid-solid polymorphic transitions (e.g., gray tin to white tin, or many polymorphic transitions in sulfur) follow exactly the same $\Delta S = \Delta H/T$ logic, though such transitions are often sluggish and harder to observe reversibly in real time.

How does this topic connect to Trouton's Rule?

Trouton's Rule observes that $\Delta_{vap}S$ clusters around 85 J/K·mol for many "normal" liquids, because their gas phases reach a similarly disordered state near boiling. This article's formula, $\Delta_{vap}S = \Delta_{vap}H/T_b$ , is exactly what you compute for each liquid to test whether it obeys Trouton's Rule. Deviations, especially in hydrogen-bonded liquids like water, signal extra structural order in the liquid that the rule's simple assumption does not capture.

How is this used to build absolute entropies from calorimetry?

Third-Law absolute entropies are built by integrating $C_P/T\, dT$ over each phase and adding $\Delta H_{trs}/T_{trs}$ at every phase transition encountered while heating from 0 K to the temperature of interest. The fusion and vaporization terms computed here are therefore essential building blocks, inserted as discrete jumps between the smooth heat-capacity integrals for solid, liquid, and gas.