6.5 Energy of Phase Changes

Enthalpy of fusion (ΔHfus, aka heat of fusion) is the molar enthalpy change for melting: the energy absorbed per mole when a solid becomes a liquid at its melting point. For n moles melting, q = n·ΔHfus. During melting the system absorbs energy and temperature stays constant (CED 6.5.A.1). Freezing releases the same magnitude of energy so ΔHfreeze = −ΔHfus (CED 6.5.A.2). Enthalpy of vaporization (ΔHvap, aka heat of vaporization) is the molar enthalpy change for converting a liquid to a gas at its boiling point. It’s usually much larger than ΔHfus for the same substance because breaking intermolecular attractions to make a gas requires more energy. Likewise, condensation releases −ΔHvap. Use q = n·ΔHvap for isothermal vaporization/condensation problems. These distinctions (melting vs. boiling, q = n·ΔH, sign conventions, constant T during phase change) are exactly what AP Topic 6.5 tests—see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and practice problems (https://library.fiveable.me/practice/ap-chemistry) for targeted practice.

When you melt a pure substance the added heat doesn’t raise temperature because it’s used to change the substance’s potential energy—breaking intermolecular attractions—not to increase kinetic energy. Temperature measures average kinetic energy of particles; during the solid→liquid transition the system absorbs heat (q = n·ΔHfus) to overcome the lattice or cohesive forces. As long as some solid remains, energy goes into changing phase (latent heat), so average kinetic energy—and thus temperature—stays constant (isothermal phase change). This is exactly what the CED says in 6.5.A.1: energy must be transferred to melt a substance, and the temperature of a pure substance remains constant during the phase change. For worked examples and to practice applying q = nΔHfus on AP-style problems, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and our practice set (https://library.fiveable.me/practice/ap-chemistry).

Use q = n · ΔHfus. Convert mass → moles, then multiply by the molar enthalpy of fusion for H2O. 1. Moles of H2O: n = 25.0 g ÷ 18.015 g·mol⁻¹ = 1.387 mol 2. Use ΔHfus (H2O) ≈ 6.01 kJ·mol⁻¹ (energy absorbed to melt 1 mol ice at 0°C). 3. q = (1.387 mol)(6.01 kJ·mol⁻¹) = 8.34 kJ → 8.3 kJ (3 sig figs). Note: this assumes the ice is already at 0°C. If it’s colder, you must first warm the ice to 0°C using q = m·c·ΔT (Topic 6.4), then use the fusion enthalpy. This directly maps to CED 6.5.A (use molar enthalpy of fusion and mole amounts). For a focused review, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and practice questions (https://library.fiveable.me/practice/ap-chemistry).

When water freezes, the system (the water) loses energy. Freezing is the reverse of melting: melting requires you to put energy into the system (positive q, ΔHfus > 0), so freezing releases that same amount of energy to the surroundings (q < 0, ΔHfreezing = −ΔHfus). During the phase change the temperature stays constant at 0°C for pure water, and the heat released equals (moles) × (molar enthalpy of fusion) with a negative sign for freezing (consistent with CED EK 6.5.A.1–A.2). For more practice and clear examples of using molar enthalpy in q = n·ΔHfus problems, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and Unit 6 review (https://library.fiveable.me/ap-chemistry/unit-6). If you want practice problems, Fiveable’s AP Chemistry practice set is a good place to apply this (https://library.fiveable.me/practice/ap-chemistry).

Heat of fusion (ΔHfus) is the molar enthalpy required to change 1 mol of a solid to a liquid at its melting point; heat of vaporization (ΔHvap) is the molar enthalpy required to change 1 mol of a liquid to a gas at its boiling point. Both are latent heats (energy absorbed without temperature change) and are positive for melting/boiling; the reverse processes (freezing, condensation) release the same magnitude of energy with opposite sign (ΔHcondensation = −ΔHvap). Use q = n·ΔH (or q = m·ΔH per gram) for AP calculations during an isothermal phase change. ΔHvap is usually much larger than ΔHfus because vaporization must completely overcome intermolecular attractions to separate molecules into the gas phase. For AP review and practice on these definitions and calculations, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and Unit 6 resources (https://library.fiveable.me/ap-chemistry/unit-6); practice problems are at (https://library.fiveable.me/practice/ap-chemistry).

Boiling water takes more energy than melting ice because the liquid→gas phase change requires breaking many more intermolecular attractions per mole than the solid→liquid change. On the AP CED this is Topic 6.5: the heat absorbed for a phase change = n · ΔH (molar enthalpy of fusion or vaporization). For H2O, ΔHfus ≈ 6.01 kJ·mol⁻¹ but ΔHvap ≈ 40.7 kJ·mol⁻¹, so vaporization needs ~6–7× more energy per mole. During both melting and boiling the temperature stays constant (isothermal phase change), and the energy goes into increasing the system’s enthalpy (latent heat) to overcome intermolecular forces. This matches EK 6.5.A.1–A.2 language (molar enthalpy of fusion/vaporization, latent heat). For a focused review, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and practice problems (https://library.fiveable.me/practice/ap-chemistry).

When a substance vaporizes (liquid → gas) you must put energy into the system to overcome intermolecular attractions, so the system’s enthalpy increases and ΔHvap is positive (endothermic). Condensation is the reverse (gas → liquid): molecules release energy as they form intermolecular attractions, so the system’s enthalpy decreases and ΔHcond is negative (exothermic). By CED 6.5.A.2, the magnitudes are equal and opposite: ΔHcond = −ΔHvap. During the phase change the temperature stays constant because the added/released energy changes enthalpy (breaking/forming interactions) rather than kinetic energy (CED 6.5.A.1). If you want a quick refresher tied to the AP framework, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and practice problems (https://library.fiveable.me/practice/ap-chemistry) to apply this on exam-style questions.

Use the sign that matches whether the system gains or loses energy. For phase changes use q = n·ΔHphase, where ΔHphase is the molar enthalpy for the specific transition (ΔHfus, ΔHvap, ΔHsub). By CED 6.5.A.1: melting and vaporization require energy input → q is positive (system absorbs heat, endothermic). Freezing and condensation release energy → q is negative (system loses heat, exothermic). Also remember: the molar enthalpy for the reverse process is the negative of the forward one (ΔHcond = −ΔHvap, ΔHfreeze = −ΔHfus). Temperature stays constant during the phase change, and always calculate n in moles before multiplying. Example: melting 2.0 mol ice: q = (2.0 mol)(+ΔHfus); freezing 2.0 mol water: q = (2.0 mol)(−ΔHfus). For the AP exam they expect you to set up q = n·ΔH correctly and give the correct sign (see Topic 6.5 study guide for examples) (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh). For extra practice, use Fiveable’s AP Chem practice problems (https://library.fiveable.me/practice/ap-chemistry).

During an isothermal phase change the temperature stays constant, so the average kinetic energy of the molecules does not change. What the added (or released) heat does is change the potential energy: it breaks (or forms) intermolecular attractions. For melting or vaporization you supply energy equal to n·ΔH(fus or vap) (latent heat) to increase the system’s internal energy as potential energy while kinetic energy (and temperature) remains constant (CED 6.5.A.1). The reverse (freezing/condensation) releases that same amount (CED 6.5.A.2). This is why phase changes are called isothermal even though energy transfers occur. For a quick review, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh); for extra practice, check the AP Chem practice set (https://library.fiveable.me/practice/ap-chemistry).

Good question—you need molar enthalpy because phase changes happen isothermally, so heat doesn’t change temperature. Heat capacity (q = m·c·ΔT or q = C·ΔT) tells you how much heat changes temperature; during melting or boiling ΔT = 0, so that formula gives q = 0—but we know energy is still absorbed or released. For phase changes you use molar enthalpy (latent heat): q = n·ΔHfus (or n·ΔHvap). That matches AP Essential Knowledge 6.5.A.1–A.2: energy is transferred at constant T and the heat required is proportional to moles and the molar enthalpy (and condensation is just the negative of vaporization). For practice problems and worked examples on this, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and more practice questions at (https://library.fiveable.me/practice/ap-chemistry).

Do it in steps (heat the ice to 0°C, melt, heat liquid to 100°C, vaporize, heat steam to 110°C). For 1.00 mol H2O (18.02 g) the calculations use Cice = 2.09 J/g·°C, Cliquid = 4.18 J/g·°C, Cgas = 2.01 J/g·°C, ΔHfus = 6.01 kJ/mol, ΔHvap = 40.7 kJ/mol. 1) Warm ice −10 → 0°C: q1 = m·Cice·ΔT = 18.02·2.09·10 = 376 J 2) Melt at 0°C: q2 = n·ΔHfus = 1.00·6.01 kJ = 6010 J 3) Heat liquid 0 → 100°C: q3 = 18.02·4.18·100 = 7530 J 4) Vaporize at 100°C: q4 = n·ΔHvap = 1.00·40.7 kJ = 40,700 J 5) Heat steam 100 →110°C: q5 = 18.02·2.01·10 = 362 J Total q = q1+q2+q3+q4+q5 = 376 + 6010 + 7530 + 40,700 + 362 = 55,0 - round = 55,0 - correct sum = 55, - let me add: 376+6010=6386; +7530=13,916; +40,700=54,616; +362=54,978 J ≈ 5.50×10^4 J (≈55.0 kJ). So for 1.00 mol H2O, total energy ≈ 5.50×10^4 J (55.0 kJ). On the AP, show each step, units, and use molar enthalpies for phase changes (CED 6.5.A). For more practice and worked examples see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and more practice problems (https://library.fiveable.me/practice/ap-chemistry).

When you add salt to ice you’re changing the liquid’s composition, which lowers the liquid’s vapor pressure and the chemical potential of the liquid phase. For a pure substance the temperature stays constant during a phase change (heat q goes into the enthalpy of fusion, ΔHfus). With dissolved salt the solid–liquid equilibrium shifts so ice and liquid water are no longer in equilibrium at 0.0 °C—the system must get colder before ice is stable. Practically, this is freezing-point depression, a colligative effect: ΔTf = i·Kf·m (for water Kf = 1.86 °C·kg/mol and NaCl roughly gives i ≈ 2). More dissolved particles (higher molality) → larger ΔTf → lower melting/freezing temperature. This ties to CED keywords: molar enthalpy of fusion, melting point, vapor pressure, and latent heat (Topic 6.5). For a quick review see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and grab practice problems at (https://library.fiveable.me/practice/ap-chemistry).

Intermolecular forces (IMFs) set how much energy you must add or remove for a phase change. Stronger IMFs (like hydrogen bonding or strong dipole–dipole) hold particles together more tightly, so you need more energy to separate them—meaning larger molar enthalpies of fusion or vaporization. Conversely, weak IMFs (like London dispersion in nonpolar gases) need less energy to overcome. AP-specific points to remember (from CED Topic 6.5): - q for a phase change = n · ΔH (use molar enthalpy of fusion, vaporization, or sublimation and moles of substance). - Temperature stays constant during an isothermal phase change. - ΔH for the reverse process has the same magnitude but opposite sign (ΔHcond = −ΔHvap, etc.). - Stronger IMFs → larger |ΔHfus| and |ΔHvap|; this also raises melting/boiling points and lowers vapor pressure. The Clausius–Clapeyron relation links ΔHvap to how vapor pressure changes with T. For a quick review, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and try practice problems (https://library.fiveable.me/practice/ap-chemistry) to apply q = n·ΔH on the exam.

Think of phase changes as either breaking or forming intermolecular “bonds.” Breaking interactions takes energy (endothermic, q absorbed); forming interactions gives off energy (exothermic, q released). So melting (solid → liquid) and vaporization (liquid → gas) absorb energy; freezing and condensation release energy. The temperature stays constant during the phase change. Use molar enthalpies: ΔHfus and ΔHvap are positive for melting/vaporizing; the reverse processes have equal magnitude but opposite sign (ΔHcond = −ΔHvap, ΔHfreeze = −ΔHfus). For calculations remember q = n · ΔH (use moles and the molar enthalpy). Keywords: latent heat, molar enthalpy of fusion/vaporization, isothermal phase change. For an AP-aligned refresher see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and try practice problems (https://library.fiveable.me/practice/ap-chemistry).

Short answer: Yes—in magnitude they’re equal but opposite in sign, provided you’re talking about the same amount of water undergoing the reverse process under the same conditions (same temperature and pressure). This is exactly what CED 6.5.A.2 states: the molar enthalpy of condensation = −(molar enthalpy of vaporization). So if vaporizing 1.00 mol H2O at its boiling point absorbs +ΔHvap, then condensing 1.00 mol steam at that same temperature/pressure releases −ΔHvap. Quick caveats: real experiments may show small differences because of heat losses to the surroundings, non-ideal conditions (superheated steam, pressure changes), or incomplete thermal equilibration—but thermodynamically the magnitudes match. For review of this topic and AP-style practice, see the Topic 6.5 study guide (https://library.fiveable.me/ap-chemistry/unit-6/energy-phase-changes/study-guide/kAXAzHrD24XL6LdpMFHh) and the Unit 6 overview (https://library.fiveable.me/ap-chemistry/unit-6).