AP CHEMISTRY • THERMOCHEMISTRY

Energy of Phase Changes

Understanding the quantitative relationships between heat transfer and transitions among solid, liquid, and gas phases.

Historical Context & Motivation

The study of heat and its relationship to phase transitions stretches back centuries, rooted in early attempts to understand why ice melts at a predictable temperature and why boiling water remains at 100 °C despite continuous heating. These observations puzzled natural philosophers who lacked a quantitative framework for distinguishing temperature from heat. The realization that energy could be absorbed or released without changing a substance's temperature was one of the most profound insights in the history of thermochemistry, eventually leading to the modern concepts of latent heat and enthalpy of phase transitions.

1761
Black's Latent Heat
Joseph Black distinguished between heat and temperature, coining the term latent heat to describe the hidden energy absorbed during melting and vaporization without a temperature change.
1780
Lavoisier & Laplace's Ice Calorimeter
Lavoisier and Laplace designed the first ice calorimeter, enabling precise measurement of heat released during chemical reactions by quantifying the mass of ice melted.
1840
Hess's Law
Germain Hess established that the total enthalpy change for a reaction is independent of the pathway taken, providing a foundation for calculating energy changes in multi-step phase transitions and chemical processes.
1882
Gibbs Free Energy
J. Willard Gibbs unified enthalpy and entropy into a single criterion for spontaneity, clarifying why phase changes occur at specific temperatures and pressures.
1930s
Modern Calorimetry
Advances in differential scanning calorimetry (DSC) allowed chemists to measure enthalpies of fusion, vaporization, and sublimation with high precision across a wide range of substances.

The central question that drove these developments remains at the heart of AP Chemistry: when a substance changes phase, how much energy is transferred, and where does that energy go if the temperature does not change? Answering this question requires understanding intermolecular forces, the distinction between kinetic and potential energy at the molecular level, and the quantitative tools of calorimetry.

Core Principles & Definitions

Phase changes occur when a substance transitions between solid, liquid, and gas states. During these transitions, energy is absorbed or released to overcome or establish intermolecular forces (IMFs) rather than to increase molecular kinetic energy. This is why temperature remains constant during a phase change at constant pressure—the added or removed energy alters the potential energy of the system, not the average kinetic energy of the particles. Understanding this distinction is essential for mastering thermochemistry on the AP Chemistry exam.

1

Enthalpy of Fusion (ΔH_fus)

The energy required to convert one mole of a solid to a liquid at its melting point. For water, ΔHfus = 6.01 kJ/mol. The reverse process (freezing) releases the same magnitude of energy.
2

Enthalpy of Vaporization (ΔH_vap)

The energy required to convert one mole of a liquid to a gas at its boiling point. For water, ΔHvap = 40.7 kJ/mol—significantly larger than ΔHfus because all IMFs must be overcome.
3

Enthalpy of Sublimation (ΔH_sub)

The energy required to convert one mole of a solid directly to a gas. By Hess's law, ΔHsub = ΔHfus + ΔHvap.
4

Endothermic vs. Exothermic

Melting, vaporization, and sublimation are endothermic (ΔH > 0); freezing, condensation, and deposition are exothermic (ΔH < 0). The sign convention reflects energy flow relative to the system.
5

Temperature Plateau

During a phase change at constant pressure, temperature remains constant because absorbed energy increases molecular potential energy (overcoming IMFs) rather than kinetic energy. This produces characteristic flat regions on a heating curve.
KEY TAKEAWAY
KEY TAKEAWAY

The Heating Curve: A Visual Explanation

A heating curve is the signature diagram of phase-change energetics. It plots temperature on the vertical axis against heat added (or time at constant heating rate) on the horizontal axis. The curve for water, the most commonly tested substance on the AP exam, reveals five distinct regions: three sloped segments where a single phase is warming and two flat plateaus where phase changes occur. The relative lengths of the plateaus directly reflect the magnitudes of ΔHfus and ΔHvap.

The heating curve for one mole of water at 1 atm. The three sloped regions represent single-phase warming (q = mcΔT), while the two flat plateaus represent phase changes (q = nΔH). Note that the boiling plateau is much wider than the melting plateau, reflecting the much larger enthalpy of vaporization.

Each sloped region in the diagram above corresponds to heating a single phase, where the heat added is calculated using q = mcΔT. The slope of each segment depends on the specific heat capacity of that phase: ice (cs = 2.09 J/(g·°C)), liquid water (cl = 4.18 J/(g·°C)), and steam (cg = 2.01 J/(g·°C)). A steeper slope means less heat is needed per degree of temperature change, corresponding to a smaller specific heat capacity. The flat plateaus, by contrast, represent regions where all added energy goes into overcoming intermolecular forces rather than raising the temperature.

Mathematical Framework

Quantitative problems involving phase changes require two distinct equations, applied to different regions of the heating curve. The key is recognizing which equation applies: within a single phase, use q = mcΔT; during a phase change, use q = nΔH. For a process that spans multiple regions (e.g., heating ice from −20 °C to steam at 120 °C), the total heat is the sum of the heat for each individual segment.

SINGLE-PHASE HEATING
q = m × c × ΔT
q = heat absorbed or released (J or kJ), m = mass (g), c = specific heat capacity (J/(g·°C)), ΔT = Tfinal − Tinitial (°C). A positive q indicates endothermic heating; a negative q indicates exothermic cooling.
PHASE CHANGE
q = n × ΔH
n = number of moles, ΔH = molar enthalpy of the phase change (kJ/mol). For fusion: ΔHfus; for vaporization: ΔHvap. Alternatively, q = m × ΔH (using mass-based enthalpy in kJ/g).
TOTAL HEAT (MULTI-SEGMENT)
q_total = q₁ + q₂ + q₃ + q₄ + q₅
For a complete heating curve with five segments: q₁ = warming solid, q₂ = melting, q₃ = warming liquid, q₄ = boiling, q₅ = warming gas. Each q is calculated with the appropriate formula for its region.
HESS'S LAW FOR SUBLIMATION
ΔH_sub = ΔH_fus + ΔH_vap
Because enthalpy is a state function, the energy to convert a solid directly to a gas equals the sum of the energies for melting and then vaporizing. This is a direct application of Hess's law to phase transitions.
AP Exam Tip

Intermolecular Forces & Enthalpy Magnitudes

The magnitude of a substance's enthalpy of phase change is directly related to the strength and type of its intermolecular forces. Substances with strong IMFs—such as hydrogen bonds, strong dipole-dipole interactions, or ionic interactions—require more energy to undergo phase transitions. This explains why water, with its extensive hydrogen-bonding network, has an unusually high ΔHvap compared to similarly sized nonpolar molecules. Conversely, substances held together only by weak London dispersion forces (such as noble gases or small nonpolar molecules) have very low enthalpies of vaporization and boil at low temperatures.

Comparison of enthalpies of vaporization across substances with different dominant intermolecular forces. Substances with hydrogen bonding (H₂O, CH₃OH) have significantly higher ΔHvap values than those with only London dispersion forces (He, N₂). NaCl, an ionic compound, has an extremely high enthalpy of vaporization (shown truncated on this scale).
Selected enthalpies of phase change and dominant intermolecular forces
SubstanceDominant IMFΔH_fus (kJ/mol)ΔH_vap (kJ/mol)Boiling Point (°C)
HeLondon dispersion0.020.08−269
N₂London dispersion0.725.6−196
CH₃OHHydrogen bonding3.1835.264.7
H₂OHydrogen bonding6.0140.7100
NaClIonic bonding28.0≈1711413

Two important trends emerge from these data. First, for every substance, ΔHvap is always substantially larger than ΔHfus. This makes physical sense: melting merely loosens the rigid structure of a solid while preserving most intermolecular contacts, whereas vaporization requires completely separating molecules from one another. Second, within each IMF category, larger or more polarizable molecules tend to have higher enthalpies of phase change due to stronger London dispersion forces contributing alongside other IMF types.

Worked Example: Full Heating Curve Calculation

Calculate the total energy required to convert 36.0 g of ice at −15.0 °C to steam at 125.0 °C at 1 atm pressure. Use the following data: cice = 2.09 J/(g·°C), cwater = 4.18 J/(g·°C), csteam = 2.01 J/(g·°C), ΔHfus = 6.01 kJ/mol, ΔHvap = 40.7 kJ/mol, MH₂O = 18.02 g/mol.

1
Step 1 — Find moles of watern = m / M = 36.0 g ÷ 18.02 g/mol = 2.00 mol. This value will be needed for the phase-change calculations in Steps 3 and 5.
n = 2.00 mol
2
Step 2 — Heat ice from −15.0 °C to 0 °Cq₁ = m × cice × ΔT = 36.0 g × 2.09 J/(g·°C) × (0 − (−15.0)) °C = 36.0 × 2.09 × 15.0 = 1,129 J = 1.13 kJ.
q₁ = 1.13 kJ
3
Step 3 — Melt ice at 0 °Cq₂ = n × ΔHfus = 2.00 mol × 6.01 kJ/mol = 12.02 kJ. Temperature remains at 0 °C throughout this step.
q₂ = 12.02 kJ
4
Step 4 — Heat liquid water from 0 °C to 100 °Cq₃ = m × cwater × ΔT = 36.0 g × 4.18 J/(g·°C) × 100.0 °C = 15,048 J = 15.05 kJ.
q₃ = 15.05 kJ
5
Step 5 — Boil water at 100 °Cq₄ = n × ΔHvap = 2.00 mol × 40.7 kJ/mol = 81.4 kJ. This is by far the largest contribution—note that boiling requires nearly six times the energy of all other steps combined.
q₄ = 81.4 kJ
6
Step 6 — Heat steam from 100 °C to 125 °Cq₅ = m × csteam × ΔT = 36.0 g × 2.01 J/(g·°C) × 25.0 °C = 1,809 J = 1.81 kJ.
q₅ = 1.81 kJ
7
Step 7 — Sum all segmentsqtotal = 1.13 + 12.02 + 15.05 + 81.4 + 1.81 = 111.4 kJ. To three significant figures, the total energy required is approximately 111 kJ.
q_total ≈ 111 kJ
Insight

Common Pitfalls & Exam Strategies

Students frequently lose points on AP Chemistry free-response questions involving phase changes due to a handful of recurring errors. Awareness of these pitfalls can make the difference between a 4 and a 5 on the exam. The table below contrasts correct reasoning with common mistakes.

Common errors in phase change calculations and their corrections
Common MistakeCorrect ApproachWhy It Matters
Using q = mcΔT during a phase changeUse q = nΔH (or q = mΔH in g-based units) when temperature is constantΔT = 0 during a phase change, so q = mcΔT gives q = 0, which is incorrect
Mixing J and kJ in multi-step calculationsConvert all values to the same unit (typically kJ) before summingA factor-of-1000 error will make the final answer wildly incorrect
Using mass (g) with molar ΔH (kJ/mol)Convert mass to moles first: n = m/M, then q = nΔHDimensional analysis fails; answer has wrong magnitude
Forgetting to use the correct specific heat for each phaseUse c_ice for solid, c_water for liquid, c_steam for gasSpecific heats differ significantly; using c_water for ice gives ≈2× error
Ignoring sign conventions for exothermic processesFreezing/condensation release heat: q < 0. Use −ΔH for reverse transitionsSign errors affect calorimetry problems and Hess's law applications
KEY TAKEAWAY
EXAM STRATEGY

Connections to Gibbs Free Energy & Phase Diagrams

The energetics of phase changes connect to deeper thermodynamic principles that extend beyond the AP Chemistry curriculum but are worth understanding conceptually. At a phase transition, the system is at equilibrium between two phases, which means ΔG = 0. Since ΔG = ΔH − TΔS, this condition gives us ΔH = TΔS at the transition temperature, or equivalently, Ttransition = ΔH/ΔS. This elegant relationship explains why each substance has a characteristic melting point and boiling point: these are the temperatures at which the enthalpy cost of disrupting intermolecular forces is exactly compensated by the entropy gain of increased molecular disorder.

Phase change concepts at AP vs. advanced levels
ConceptAP Chemistry LevelAdvanced / College Level
Why do phase changes occur?Energy input overcomes IMFs; energy output establishes IMFsΔG = 0 at the transition temperature; competing ΔH and TΔS terms drive spontaneity
Effect of pressureQualitative: higher pressure raises boiling pointClausius-Clapeyron equation: ln(P₂/P₁) = (ΔH_vap/R)(1/T₁ − 1/T₂)
Phase diagramsIdentify regions (solid, liquid, gas), triple point, critical pointPhase boundaries derived from dP/dT = ΔS/ΔV (Clapeyron equation); supercritical fluids
Enthalpy calculationsq = nΔH and q = mcΔT for each segmentTemperature-dependent ΔH using Kirchhoff's equation: ΔH(T₂) = ΔH(T₁) + ∫ΔC_p dT

For the AP exam, you are expected to perform multi-step heating curve calculations, interpret heating/cooling curves qualitatively, and connect the magnitude of phase-change enthalpies to the strength of intermolecular forces. The Clausius-Clapeyron equation and Gibbs free energy derivations are beyond the scope of the AP curriculum, but understanding the conceptual relationship between ΔH, ΔS, and the transition temperature provides valuable insight when answering qualitative free-response questions about why certain substances have higher or lower melting/boiling points.

Practice Problems

1
A beaker of water is heated at a constant rate. During the time the water is boiling, the temperature of the water remains constant at 100 °C. Which of the following best explains why the temperature does not increase during boiling?
2
How much energy is required to melt 90.0 g of ice at 0 °C? (ΔHfus for water = 6.01 kJ/mol; molar mass of H₂O = 18.02 g/mol)
3
A 50.0 g sample of liquid water at 25.0 °C is heated until it becomes steam at 100.0 °C. Which of the following correctly describes the energy contributions? (cwater = 4.18 J/(g·°C), ΔHvap = 40.7 kJ/mol, M = 18.02 g/mol)
PROBLEM 4APPLIED
A student places a 150.0 g block of ice at −10.0 °C into an insulated container holding 400.0 g of liquid water at 60.0 °C. Assume no heat is lost to the surroundings. Data: cice = 2.09 J/(g·°C), cwater = 4.18 J/(g·°C), ΔHfus = 334 J/g. (a) Calculate the energy, in joules, needed to warm the ice from −10.0 °C to 0 °C. (b) Calculate the energy, in joules, needed to melt all the ice at 0 °C. (c) Determine whether the warm water can supply enough energy to melt all the ice. Set up the equation qlost + qgained = 0 and solve for the final equilibrium temperature of the system. (d) Explain, at the molecular level, why the final temperature of the system is significantly lower than the initial temperature of the warm water.
PROBLEM 5CRITICAL THINKING
A researcher measures the enthalpy of vaporization for four liquids and records the following data: (a) Substance A is a nonpolar gas at room temperature. Using the data, explain why its ΔH_vap is the lowest of the four substances. (b) Substance D has the smallest molar mass but the largest ΔH_vap. Identify the dominant IMF in Substance D, and explain why molar mass alone does not predict ΔH_vap. (c) Rank substances B, C, and D in order of increasing boiling point. Justify your ranking using the data. (d) Predict how the ΔH_sub of Substance D compares to its ΔH_vap. Justify your answer using Hess's law. (e) If Substance C is ethanol (C₂H₅OH), a student claims that its ΔH_vap should be higher than that of Substance D because ethanol has a larger molar mass. Evaluate this claim.
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