Physical Chemistry 1 • Thermodynamic Foundations

Entropy Changes: Phase Changes — Compute entropy changes for phase changes and idealized paths

Master the thermodynamic bookkeeping of disorder through reversible phase transitions and constructed state-function paths.

An atom with a chemistry flask and thermometer for physical chemistry fundamentals.
Inside this lesson 10 sections 5 questions 8 figures

SECTION 1

Historical Context & Motivation

The concept of entropy arose from efforts to understand the fundamental limits of heat engines and the irreversibility of natural processes. In the early nineteenth century, engineers and physicists recognized that heat could not be converted entirely into work, and this realization demanded a new thermodynamic quantity to describe the 'lost' capacity for useful transformation. Phase changes—melting, boiling, sublimation—offered some of the cleanest experimental windows into this new quantity, because they occur at constant temperature and pressure, simplifying the mathematical treatment considerably. Understanding how entropy changes during these transitions became a cornerstone of classical thermodynamics and remains essential for predicting equilibrium, designing industrial processes, and connecting macroscopic measurements to molecular-level disorder.

1824 • Carnot's Reflections

Sadi Carnot publishes Réflexions sur la puissance motrice du feu, establishing that the efficiency of heat engines depends only on the temperatures of the hot and cold reservoirs—an insight that implicitly contains the seeds of entropy.

1850 • Clausius Formalizes the Second Law

Rudolf Clausius articulates the second law of thermodynamics and introduces the concept that heat cannot spontaneously flow from cold to hot bodies, laying the groundwork for defining entropy as a state function.

1865 • Entropy Named

Clausius coins the term Entropie (from the Greek τροπή, meaning 'transformation') and writes the famous inequality δq/T ≤ dS, with equality holding for reversible processes. This formalization enables direct calculation of entropy changes for phase transitions.

1877 • Boltzmann's Statistical Interpretation

Ludwig Boltzmann connects entropy to the number of microstates via S = kB ln Ω, providing a molecular rationale for why entropy increases during melting and vaporization—more accessible configurations become available.

1923 • Lewis and Randall Tabulate Standard Entropies

Gilbert N. Lewis and Merle Randall publish comprehensive tables of standard molar entropies using calorimetric data on heat capacities and phase-transition enthalpies, making practical entropy calculations routine for chemists.

The central question that motivated these developments is deceptively simple: how do we quantify the entropy change when a substance undergoes a phase transition, and how do we extend that calculation to processes that follow non-trivial paths through state space? Because entropy is a state function, we are free to devise any idealized reversible path connecting the initial and final states, compute ΔS along that path, and the result will be valid for any real process between the same endpoints. This powerful strategy is the focus of this lesson.

SECTION 2

Core Principles & Definitions

Before computing entropy changes for phase transitions, several foundational ideas must be firmly in place. The overarching principle is that entropy is a state function: its value depends only on the current thermodynamic state (T, P, phase, composition), not on the path by which that state was reached. This path-independence is what licenses us to replace a messy irreversible process with a convenient reversible one for calculation purposes. The Clausius definition provides the operational recipe: for a reversible process, the infinitesimal entropy change is dS = δqrev / T. When integrated over a finite process, ΔS = ∫ δqrev / T, and evaluating this integral under various constraints yields all the expressions used in this lesson.

1

State Function Property

ΔS between two states is path-independent. Any reversible path connecting the same initial and final states yields the same ΔS, even if the real process is irreversible.
2

Reversible Phase Change at T and P

At the equilibrium transition temperature (e.g., normal boiling point), the phase change is reversible. The entropy change is simply ΔS = ΔtrsH / Ttrs, since T is constant throughout the transition.
3

Heat Capacity Integration

For heating or cooling within a single phase (no transition), ΔS = ∫ Cp dT / T. When Cp is approximately constant, this simplifies to Cp ln(T₂/T₁).
4

Constructing Idealized Paths

When a phase change occurs at a non-equilibrium temperature (e.g., supercooled water freezing at −10 °C), we construct a multi-step reversible path: heat to the equilibrium T, carry out the reversible transition, then cool to the actual final T.
5

Trouton's Rule

For many non-polar, non-hydrogen-bonded liquids, the standard molar entropy of vaporization at the normal boiling point is approximately 85 J mol⁻¹ K⁻¹. Significant deviations indicate strong intermolecular interactions (e.g., water ≈ 109 J mol⁻¹ K⁻¹).
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KEY TAKEAWAY

Think of entropy calculations like computing the elevation change between two cities. No matter which highway (path) you drive, the altitude difference is the same because elevation is a state function of position—just as entropy is a state function of thermodynamic state. When the direct road is winding and unmapped (irreversible), you simply choose a well-surveyed route (reversible path) and read off the altitude change from the surveyor's data.

SECTION 3

Visual Explanation: Entropy Along a Heating Curve

Entropy vs. Temperature for a Pure Substance at Constant PressureTemperature (T) →Entropy (S) →Solid∫Cp,sdT/TΔfusS= ΔfusH / TfusTfusLiquid∫Cp,ldT/TΔvapS= ΔvapH / TvapTvapGas∫Cp,gdT/T
The S-vs-T diagram shows entropy increasing continuously within each phase (solid, liquid, gas) according to Cp dT/T integration. At each phase boundary (Tfus and Tvap), vertical jumps correspond to the isothermal absorption of the latent heat (ΔtrsH / Ttrs). The vaporization jump is always much larger than the fusion jump because ΔvapH ≫ ΔfusH for virtually all substances.

The diagram above encodes the entire strategy for computing ΔS between any two states that span one or more phase boundaries. Starting from a temperature in the solid phase and ending in the gas phase, for example, the total entropy change is the sum of five additive contributions: heating the solid to Tfus, the isothermal entropy of fusion, heating the liquid from Tfus to Tvap, the isothermal entropy of vaporization, and finally heating the gas from Tvap to the final temperature. Each of these segments is evaluated independently and then summed, exploiting the additivity of entropy as a state function.

SECTION 4

Mathematical Framework

We now derive the key expressions used throughout this lesson. All derivations start from the Clausius equality for reversible processes: dS = δqrev / T. For an isobaric (constant-pressure) process, δqrev = dH = Cp dT (within a phase) or δqrev = ΔtrsH (at a phase boundary).

ENTROPY OF A REVERSIBLE PHASE TRANSITION

ΔtrsS=ΔtrsH/TtrsΔtrsS = ΔtrsH / Ttrs

ΔtrsH = molar enthalpy of the transition (J mol⁻¹); Ttrs = equilibrium transition temperature (K). At constant pressure, this is exact because the transition is isothermal and reversible when performed at the equilibrium T.

ENTROPY CHANGE FOR HEATING/COOLING WITHIN A PHASE

ΔS=∫(T1→T2)CpdT/T=Cpln(T2/T1)[Cpconstant]ΔS = ∫(T₁→T₂) Cp dT / T = Cp ln(T₂ / T₁) [Cp constant]

Cp = isobaric molar heat capacity (J mol⁻¹ K⁻¹). If Cp varies with temperature, retain the integral and use the functional form Cp(T) = a + bT + cT⁻² or equivalent.

COMPOSITE PATH SPANNING MULTIPLE PHASES (solid → gas)

ΔS=Cp,sln(Tfus/T1)+ΔfusH/Tfus+Cp,lln(Tvap/Tfus)+ΔvapH/Tvap+Cp,gln(T2/Tvap)ΔS = Cp,s ln(Tfus/T₁) + ΔfusH/Tfus + Cp,l ln(Tvap/Tfus) + ΔvapH/Tvap + Cp,g ln(T₂/Tvap)

Each additive term corresponds to one segment of the S-vs-T diagram: heating the solid, isothermal fusion, heating the liquid, isothermal vaporization, heating the gas. Subscripts s, l, g denote solid, liquid, and gas heat capacities, respectively.

IDEALIZED PATH FOR AN IRREVERSIBLE PHASE CHANGE

ΔS=Cp(phase1)ln(Ttrs/T)+ΔtrsH/Ttrs+Cp(phase2)ln(T/Ttrs)ΔS = Cp(phase 1) ln(Ttrs/T) + ΔtrsH/Ttrs + Cp(phase 2) ln(T/Ttrs)

When a phase change occurs at a temperature T that is not the equilibrium transition temperature (e.g., supercooled water freezing at −10 °C), construct a three-step reversible path: (1) bring the substance from T to Ttrs in phase 1, (2) carry out the reversible phase transition at Ttrs, (3) bring the substance from Ttrs back to T in phase 2. Sum the three ΔS contributions.

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Sign Convention Note

For exothermic transitions (e.g., freezing, condensation), ΔtrsH is negative, so ΔtrsS is also negative—the system becomes more ordered. Always check that your sign is physically reasonable: does the substance gain or lose microscopic freedom?

SECTION 5

Constructing Idealized Reversible Paths

The real power of the state-function property of entropy is unleashed when a phase change occurs under conditions that are irreversible. Consider a classic example: liquid water supercooled to −10 °C (263.15 K) at 1 atm suddenly crystallizes. This process is irreversible—it happens spontaneously and cannot be reversed by an infinitesimal change in conditions. Clausius's definition dS = δqrev / T cannot be applied directly to the actual irreversible path. Instead, we construct a hypothetical multi-step reversible path connecting the same initial state (liquid at 263.15 K) and final state (ice at 263.15 K). Because ΔS is path-independent, the entropy change computed along this hypothetical path is exactly the entropy change of the irreversible process.

Idealized Reversible Path for Irreversible Freezing(Supercooled water at 263 K → Ice at 263 K)Initial StateH₂O(l), 263.15 KFinal StateH₂O(s), 263.15 KIrreversible (actual)ΔS = ? (cannot integrate directly)Step 1Heat liquid263.15 K → 273.15 KΔS₁ = Cp,l ln(273.15/263.15)Step 2Freeze at 273.15 K(reversible, isothermal)ΔS₂ = −ΔfusH / 273.15Step 3Cool ice273.15 K → 263.15 KΔS₃ = Cp,s ln(263.15/273.15)Total: ΔS = ΔS₁ + ΔS₂ + ΔS₃Same ΔS as the irreversible process (state function!)
The three-step idealized path replaces the irreversible freezing of supercooled water with a sequence of reversible steps: heat the liquid to 273.15 K, freeze reversibly at 273.15 K, and cool the ice back to 263.15 K. Each step's ΔS is calculable from standard formulas, and their sum equals the entropy change of the actual irreversible process.

This strategy generalizes immediately. Whenever a real process involves a phase change at a temperature other than the equilibrium transition temperature, or involves heating across multiple phase boundaries, you decompose the path into segments where the entropy integral can be evaluated analytically. The only data you need are the heat capacities of each phase and the enthalpies of transition at the equilibrium temperatures. This approach is used extensively in materials science, chemical engineering, and geochemistry to compute absolute entropies from 0 K calorimetric data using the Third Law.

SECTION 6

Worked Example: Freezing of Supercooled Water

Calculate the molar entropy change of the system when supercooled liquid water at −10 °C (263.15 K) freezes irreversibly to ice at −10 °C at 1 atm. Use the following data: ΔfusH = 6.01 kJ mol⁻¹ at 273.15 K, Cp(liquid) = 75.3 J mol⁻¹ K⁻¹, Cp(ice) = 38.0 J mol⁻¹ K⁻¹. Assume heat capacities are independent of temperature over this range.

Molar Entropy Change for Irreversible Freezing of Supercooled Water

  1. 1

    Step 1 — Identify the Initial and Final States

    Initial state: H₂O(l) at T = 263.15 K, P = 1 atm. Final state: H₂O(s) at T = 263.15 K, P = 1 atm. The process is irreversible—supercooled water spontaneously crystallizes. We cannot apply dS = δqrev / T directly, so we construct a reversible path.
  2. 2

    Step 2 — Construct the Idealized Reversible Path

    Path: (a) Reversibly heat the liquid from 263.15 K to 273.15 K. (b) Reversibly freeze the liquid at 273.15 K (the equilibrium freezing point). (c) Reversibly cool the ice from 273.15 K back to 263.15 K.
  3. 3

    Step 3 — Compute ΔS₁ (Heating the Liquid)

    ΔS₁ = Cp,l × ln(T₂/T₁) = 75.3 J mol⁻¹ K⁻¹ × ln(273.15 / 263.15) = 75.3 × ln(1.0380) = 75.3 × 0.03731
    ΔS₁ = +2.81 J mol⁻¹ K⁻¹
  4. 4

    Step 4 — Compute ΔS₂ (Reversible Freezing at 273.15 K)

    Freezing is the reverse of fusion, so we use −ΔfusH: ΔS₂ = −ΔfusH / Tfus = −6010 J mol⁻¹ / 273.15 K
    ΔS₂ = −22.00 J mol⁻¹ K⁻¹
  5. 5

    Step 5 — Compute ΔS₃ (Cooling the Ice)

    ΔS₃ = Cp,s × ln(T₂/T₁) = 38.0 J mol⁻¹ K⁻¹ × ln(263.15 / 273.15) = 38.0 × ln(0.9634) = 38.0 × (−0.03731)
    ΔS₃ = −1.42 J mol⁻¹ K⁻¹
  6. 6

    Step 6 — Sum the Contributions

    ΔSsys = ΔS₁ + ΔS₂ + ΔS₃ = 2.81 + (−22.00) + (−1.42) = −20.61 J mol⁻¹ K⁻¹. The negative sign is physically correct: freezing reduces the system's disorder. Note that the total entropy of the universe (system + surroundings) must still be positive for this spontaneous process.
    ΔSsys = −20.6 J mol⁻¹ K⁻¹

SECTION 7

Strengths, Limitations, and Common Pitfalls

AspectStrengthLimitation / Pitfall
State-function strategyAny reversible path gives the correct ΔS, regardless of how the actual irreversible process occurs.Requires knowledge of the equilibrium transition temperature and enthalpy; if these data are unavailable, the method cannot be applied directly.
Constant Cp assumptionSimplifies the integral to Cp ln(T₂/T₁), making calculations analytically tractable.Cp varies with temperature; over wide temperature ranges (>50 K), neglecting this variation introduces significant error.
Trouton's RuleQuick estimation of ΔvapS ≈ 85 J mol⁻¹ K⁻¹ for non-associated liquids.Fails for hydrogen-bonded, metallic, or strongly polar liquids (water, ethanol, mercury).
Additivity of path segmentsComplex paths crossing multiple phase boundaries are handled by simply summing individually computed ΔS values.Missing or incorrectly ordered segments will produce erroneous results; always verify that the constructed path truly connects the initial and final states.
Sign of ΔtrsHClear physical meaning: positive for endothermic transitions (melting, vaporization), negative for exothermic ones.A common error is using a positive ΔfusH when computing entropy of freezing; always use the sign appropriate to the direction of the transition.

Strengths and limitations of the idealized-path entropy calculation method

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KEY TAKEAWAY

The idealized-path method is analogous to how GPS navigation calculates net elevation change: even if you hike off-trail through dense forest (an irreversible path), the elevation difference between trailhead and summit is the same as if you walked the paved switchback road (a reversible path). As long as you carefully account for every up and down along your chosen reversible route, your answer is exact. The most common mistake is forgetting a 'switchback'—that is, omitting one of the heating or cooling segments in the constructed path.

SECTION 8

Connection to Advanced Theory

The techniques developed in this lesson form the foundation for several more advanced thermodynamic treatments. In particular, the ability to compute ΔS along idealized paths feeds directly into the Third Law of Thermodynamics and the determination of absolute (Third-Law) entropies. By integrating Cp/T from near 0 K to any desired temperature—pausing to add ΔtrsH/Ttrs at each phase boundary—we obtain S°(T), the standard molar entropy at temperature T. Furthermore, these entropy calculations underpin the Gibbs energy criterion for spontaneity: ΔG = ΔH − TΔS, which is the central quantity in chemical equilibrium, electrochemistry, and biochemistry.

This LessonAdvanced Extension
ΔtrsS = ΔtrsH / Ttrs for a single transitionClausius–Clapeyron equation: dP/dT = ΔtrsS / ΔtrsV relates entropy of transition to the slope of the phase boundary in the P-T diagram.
Multi-segment path with constant CpTemperature-dependent Cp(T) = a + bT + c/T² used in Kirchhoff's equation and Shomate polynomials for precise Third-Law entropy calculations.
System entropy change onlyTotal entropy (ΔSuniv = ΔSsys + ΔSsurr ≥ 0) used as the criterion for spontaneity; leads to Gibbs and Helmholtz free energies.
Trouton's Rule as an estimateStatistical mechanical derivation via the Sackur–Tetrode equation for ideal gas translational entropy explains why ΔvapS is roughly constant for simple liquids.

How this lesson's concepts connect to advanced thermodynamics

Looking forward, when you study the Gibbs phase rule and phase diagrams, you will see that the entropy difference between coexisting phases is precisely what determines the slope of the phase boundary line. The Clausius–Clapeyron equation, dP/dT = ΔtrsS / ΔtrsV, directly employs the ΔtrsS values you are now learning to compute. Mastery of entropy calculations for phase changes therefore serves as the quantitative gateway to understanding all of chemical phase equilibrium.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Explain why the entropy of vaporization is always much larger in magnitude than the entropy of fusion for the same substance. Relate your answer to the molecular-level changes that occur in each transition.
Show answer
During fusion (melting), molecules transition from a rigid lattice to a disordered liquid, but they remain in close contact and retain significant short-range order. The enthalpy of fusion is relatively modest. During vaporization, molecules escape all significant intermolecular interactions and occupy a volume roughly 1000 times larger than the liquid, dramatically increasing the number of accessible microstates. The enthalpy of vaporization is correspondingly much larger—typically 5–10 times ΔfusH. Because ΔtrsS = ΔtrsH / Ttrs and the boiling point is also higher than the melting point (making the denominator larger), the dominant factor is still the much larger numerator, ensuring ΔvapS ≫ ΔfusS.

Question 2: BASIC CALCULATION

Calculate the molar entropy of vaporization of benzene at its normal boiling point (353.3 K) given ΔvapH = 30.72 kJ mol⁻¹. Does benzene obey Trouton's Rule?
Show answer
ΔvapS = ΔvapH / Tvap = 30720 J mol⁻¹ / 353.3 K = 86.9 J mol⁻¹ K⁻¹. Trouton's Rule predicts approximately 85 J mol⁻¹ K⁻¹ for non-associated liquids, and benzene's value of 86.9 J mol⁻¹ K⁻¹ falls squarely within the expected range. Benzene is non-polar and non-hydrogen-bonding, so this agreement is expected and validates the empirical rule for this class of substances.

Question 3: INTERMEDIATE

Calculate the total molar entropy change when 1 mol of ice at −25 °C is converted to steam at 125 °C at 1 atm. Data: Cp(ice) = 38.0 J mol⁻¹ K⁻¹, Cp(water) = 75.3 J mol⁻¹ K⁻¹, Cp(steam) = 36.4 J mol⁻¹ K⁻¹, ΔfusH = 6.01 kJ mol⁻¹ at 273.15 K, ΔvapH = 40.67 kJ mol⁻¹ at 373.15 K.
Show answer
This requires five segments. (1) Heat ice: 38.0 × ln(273.15/248.15) = 38.0 × 0.09609 = 3.65 J mol⁻¹ K⁻¹. (2) Fusion: 6010/273.15 = 22.00. (3) Heat water: 75.3 × ln(373.15/273.15) = 75.3 × 0.3120 = 23.49. (4) Vaporization: 40670/373.15 = 109.0. (5) Heat steam: 36.4 × ln(398.15/373.15) = 36.4 × 0.06494 = 2.36. Total ΔS = 3.65 + 22.00 + 23.49 + 109.0 + 2.36 = 160.5 J mol⁻¹ K⁻¹. Note that the vaporization step dominates, contributing nearly 68% of the total entropy increase.

Question 4: APPLIED

In a metallurgical process, liquid tin at 505.08 K (the normal melting point) solidifies and then cools to 400 K. Given ΔfusH(Sn) = 7.03 kJ mol⁻¹ and Cp(solid Sn) = 27.0 J mol⁻¹ K⁻¹, calculate the molar entropy change of the system for this process.
Show answer
Two segments: (1) Reversible freezing at 505.08 K: ΔS₁ = −ΔfusH / Tfus = −7030 / 505.08 = −13.92 J mol⁻¹ K⁻¹. (2) Cool solid tin from 505.08 K to 400 K: ΔS₂ = 27.0 × ln(400/505.08) = 27.0 × (−0.2335) = −6.30 J mol⁻¹ K⁻¹. Total ΔS = −13.92 + (−6.30) = −20.2 J mol⁻¹ K⁻¹. The negative sign confirms that the system becomes more ordered as the liquid freezes and the solid cools, consistent with the Second Law as long as the surroundings gain more entropy than the system loses.

Question 5: CRITICAL THINKING

Supercooled liquid water at 263.15 K (−10 °C) freezes irreversibly at constant pressure. Using the data from the worked example, compute both the system and surroundings entropy changes, and verify that ΔSuniv > 0. For the surroundings, assume they act as a thermostat at 263.15 K and absorb qsurr = −qsys. Calculate qsys for the actual irreversible process, then discuss what drives this process spontaneously.
Show answer
From the worked example, ΔSsys = −20.6 J mol⁻¹ K⁻¹. For qsys of the actual process (constant P), qsys = ΔH = −ΔfusH(at 263 K). Using Kirchhoff: ΔfusH(263) = 6010 + (75.3 − 38.0)(263.15 − 273.15) = 6010 − 373 = 5637 J mol⁻¹. So qsys = −5637 J mol⁻¹ (exothermic). ΔSsurr = −qsys/Tsurr = 5637/263.15 = +21.4 J mol⁻¹ K⁻¹. ΔSuniv = −20.6 + 21.4 = +0.8 J mol⁻¹ K⁻¹ > 0, confirming spontaneity. The process is driven by the entropy gain of the surroundings (which receive heat) exceeding the entropy loss of the system (which becomes more ordered). This illustrates that a decrease in system entropy is permissible as long as the surroundings' entropy increases by a larger amount.

SUMMARY

Lesson Summary

Summary

This lesson established how to compute entropy changes for phase transitions using the Clausius definition dS = δqrev / T. For a reversible phase change at the equilibrium temperature, the entropy change is simply ΔtrsS = ΔtrsH / Ttrs. For heating or cooling within a single phase, ΔS = Cp ln(T₂/T₁) when the heat capacity is constant. The key insight is that because entropy is a state function, we can construct idealized multi-step reversible paths to compute ΔS for irreversible processes—such as the freezing of supercooled water—by routing through the equilibrium transition temperature.

Practical applications include computing absolute (Third-Law) entropies from 0 K calorimetric data, estimating vaporization entropies via Trouton's Rule (≈ 85 J mol⁻¹ K⁻¹ for non-associated liquids), and providing the entropy terms needed for Gibbs energy calculations that determine spontaneity and equilibrium. These techniques form the quantitative backbone of equilibrium thermodynamics and connect directly to the Clausius–Clapeyron equation, phase diagrams, and chemical potential theory studied later in the course.

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