6.5 Energy of Phase Changes

Phase changes happen at a constant temperature, and the energy involved is calculated with $q=n\Delta H$, where $\Delta H$ is the molar enthalpy of the phase change. Melting and boiling absorb energy, while freezing and condensing release the same amount of energy. For AP Chemistry, use the sign of $q$ to show whether the process is endothermic or exothermic.

AP Chem 6.5 Energy of Phase Changes Summary

In AP Chemistry 6.5, phase-change energy depends on the amount of substance and the molar enthalpy of the phase transition: $q=n\Delta H$. For enthalpy values given per gram, use mass instead: $q=m\Delta H$.

During a phase change, the temperature of a pure substance stays constant. Melting and vaporization absorb energy, while freezing and condensation release energy with the same magnitude and the opposite sign.

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Why This Matters for the AP Chemistry Exam

This topic connects particle-level behavior to measurable energy changes, which is exactly the kind of reasoning AP Chemistry rewards. You will need to describe what is happening to a substance during a phase change and pull quantitative information out of models like heating and cooling curves.

On the exam, phase-change energy shows up in two main ways:

• Calculating heat using q = n·ΔH (or q = m·ΔH if the enthalpy is given per gram) for a melting, freezing, boiling, or condensing step.
• Reading heating and cooling curves to identify where temperature changes versus where a phase transition is happening, and connecting that to whether energy is absorbed or released.

These skills combine with calorimetry from earlier in Unit 6, so multi-step problems often ask you to track energy across both temperature changes and phase changes.

Key Takeaways

• The temperature of a pure substance stays constant during a phase change. All added or removed energy goes into changing the phase, not the temperature.
• Melting (fusion) and boiling (vaporization) are endothermic; freezing and condensing are exothermic.
• Complementary phase changes have equal magnitude and opposite sign: ΔH of condensation = −ΔH of vaporization, and ΔH of freezing = −ΔH of fusion.
• Use q = n·ΔH for phase-change steps and q = mcΔT for temperature-change steps. Never mix them up.
• The molar enthalpy of vaporization is larger than the molar enthalpy of fusion because vaporizing separates particles more completely, breaking more intermolecular forces.
• Watch your units. Enthalpy can be given in J/g, kJ/g, J/mol, or kJ/mol, so convert before plugging in.

Heating Curves

A heating curve shows how a substance's temperature changes as you add energy at a steady rate.

The x-axis is the amount of energy added over time, and the y-axis is temperature. As energy goes in, the substance moves from solid to liquid to gas. The sloped sections are where the temperature is rising, and the flat sections (plateaus) are where a phase change is happening.

What the Plateaus Mean

The plateaus are where things get interesting. During a plateau, you are adding energy but the temperature is not changing. That energy is going into breaking intermolecular forces to change the phase instead of speeding up the particles.

The two plateaus are the melting plateau (using the molar enthalpy of fusion, ΔH_fus) and the vaporizing plateau (using the molar enthalpy of vaporization, ΔH_vap). A solid stays at its melting point until it is completely melted, then the temperature starts rising again. The same idea applies at the boiling point.

Why Vaporizing Takes More Energy

The molar enthalpy of vaporization is almost always larger than the molar enthalpy of fusion, so the vaporizing plateau is longer. It takes more energy to boil a substance than to melt it.

Think about it in terms of intermolecular forces. When a solid melts, particles loosen but stay close together, so only some intermolecular forces are disrupted. When a liquid boils, particles separate completely into a gas, so far more intermolecular forces must be overcome. That extra work shows up as the larger enthalpy of vaporization.

Worked Example

How many joules are required to change 30.0 g of ice at -20°C to steam at 140°C?

Given information:

• Specific heat of ice: 2.108 J/g·°C
• Specific heat of water: 4.18 J/g·°C
• Specific heat of steam: 2.010 J/g·°C
• ΔH_fus for H2O = 334 J/g
• ΔH_vap for H2O = 2260 J/g

This problem has five steps. Use q = mcΔT on the slopes (temperature is changing) and q = m·ΔH on the plateaus (phase is changing).

• Solid (warming ice): q = mcΔT = (30.0 g)(2.108 J/g·°C)(20°C) = 1264.8 J. ΔT comes from -20°C to 0°C, the melting point of water.
• Melting: q = m·ΔH_fus = (30.0 g)(334 J/g) = 10020 J
• Liquid (warming water): q = mcΔT = (30.0 g)(4.18 J/g·°C)(100°C) = 12540 J. ΔT comes from 0°C to 100°C.
• Vaporizing: q = m·ΔH_vap = (30.0 g)(2260 J/g) = 67800 J
• Gas (warming steam): q = mcΔT = (30.0 g)(2.010 J/g·°C)(40°C) = 2412 J. ΔT comes from 100°C to 140°C.

Add them all together: 94,036.8 J, or about 94.0 kJ.

Cooling Curves

A cooling curve is the reverse of a heating curve. Instead of absorbing energy, the substance releases it, so the processes are exothermic.

The shape is the same as a heating curve, but the plateaus now represent condensation and freezing instead of vaporization and melting. Because complementary phase changes have equal magnitude and opposite sign, the enthalpy of condensation is the negative of the enthalpy of vaporization, and the enthalpy of freezing is the negative of the enthalpy of fusion.

Pay attention to the units you are given. In the example above, ΔH_fus and ΔH_vap were in J/g, but a problem could give you J/mol or kJ/mol. If the enthalpy is given per mole, convert the mass into moles before using q = n·ΔH.

How to Use This on the AP Chemistry Exam

Problem Solving

• Sketch or label a heating curve so you can see which steps are slopes (q = mcΔT) and which are plateaus (q = n·ΔH or q = m·ΔH). Multi-step problems become much easier when you separate them.
• Track your units before you calculate. If the enthalpy is per mole, convert grams to moles first. If it is per gram, you can use mass directly.
• For a phase change in the opposite direction, flip the sign. Freezing releases the same magnitude of energy that melting absorbs.
• Add up all the steps for the total energy. Do not forget any temperature-change segments between phase transitions.

Free Response

• When asked to explain why a temperature plateau exists, state that the added energy is breaking intermolecular forces to change phase rather than increasing the average kinetic energy of the particles.
• When comparing ΔH_vap and ΔH_fus, justify your answer using intermolecular forces: vaporization separates particles completely, so it overcomes more attractive forces and requires more energy.
• Be precise about endothermic versus exothermic. Melting and boiling absorb energy; freezing and condensing release energy.

Common Trap

Mixing up q = mcΔT and q = n·ΔH is the most frequent mistake. Remember: temperature is changing on the slopes (use specific heat) and phase is changing on the plateaus (use molar or per-gram enthalpy). The temperature does not change during a phase transition, so ΔT is zero there and q = mcΔT would give you zero.

Common Misconceptions

• "Temperature keeps rising during melting or boiling." It does not. The temperature of a pure substance stays constant during a phase change. All the energy goes into the phase transition.
• "q = mcΔT works for the whole process." It only works for the sloped sections where temperature changes. On plateaus, you must use q = n·ΔH (or q = m·ΔH).
• "Freezing and condensing absorb energy." They release energy. They are exothermic and are the reverse of melting and boiling.
• "ΔH_fus and ΔH_vap are the same size." Vaporization usually requires much more energy than fusion because boiling fully separates particles and breaks more intermolecular forces.
• "You can ignore the units on enthalpy values." You cannot. J/g and J/mol are different. Match the enthalpy units to whether you are using mass or moles before calculating.

Related AP Chemistry Guides

• Unit 6 Overview: Thermochemistry
• 6.1 Endothermic and Exothermic Processes
• 6.2 Energy Diagrams
• 6.4 Heat Capacity and Calorimetry
• 6.3 Heat Transfer and Thermal Equilibrium
• 6.6 Introduction to Enthalpy of Reaction