Enthalpy of Phase Changes
Fusion, vaporisation and sublimation as state-dependent energy changes
Lesson 1742 of 4,500 · Thermodynamics
Learning objectives
- Distinguish fusion, vaporisation and sublimation enthalpies
- Use Hess's law to link phase changes and reverse processes
Introduction
Changing a substance's phase changes its enthalpy even when its chemical formula stays the same. Melting, vaporization and sublimation usually absorb heat at the transition conditions, while freezing, condensation and deposition release it. Phase-change enthalpies explain why physical-state labels are indispensable in thermochemical equations.
Core explanation
At a stated pressure and transition temperature, the molar enthalpy of fusion ΔH fus describes solid → liquid for one mole. Vaporization ΔH vap describes liquid → gas, and sublimation ΔH sub describes solid → gas. For ordinary stable phases at their equilibrium transitions, the forward processes generally require positive enthalpy because particles move into phases with weaker average intermolecular constraints. The reverse enthalpies have equal magnitude and opposite sign at the same conditions.
For water at a specified pressure, H₂O(s) → H₂O(l) is fusion; H₂O(l) → H₂O(g) is vaporization; H₂O(s) → H₂O(g) is sublimation. Hess's law gives ΔH sub = ΔH fus + ΔH vap when the staged and direct routes connect the same initial and final states at a common temperature and pressure. If tabulated fusion and vaporization values are measured at different temperatures, heat-capacity corrections are needed before adding them precisely.
During an equilibrium phase change at fixed pressure, temperature can remain constant while heat is absorbed or released. Therefore q = mcΔT alone would give zero and miss the latent heat. A heating calculation that crosses melting must include warming the solid, then fusion enthalpy, then warming the liquid. If it later boils, add vaporization separately. Every term must refer to its own phase and temperature interval.
The distinction between H₂O(l) and H₂O(g) in combustion is a common application. If a reaction forms liquid water, its enthalpy includes the energy released as water vapor condenses to liquid at matching conditions. A reaction written with gaseous water lacks that condensation contribution. The change is not a rounding detail; it can materially alter a fuel's reported heating value.
Phase changes involve intermolecular interactions, but the simplistic phrase “bonds break during boiling” can mislead. In vaporizing water, most O–H covalent bonds inside molecules remain intact. The energy primarily changes interactions between molecules and the PV contribution of the gas phase. Sublimation of an ionic or network solid has different structural details; the specific substance must be identified.
Pressure affects transition temperatures and phase equilibria. A molar vaporization enthalpy at one temperature is not automatically identical at another. Near a critical point, the distinction between liquid and gas diminishes. Introductory calculations often treat ΔH values as constant over a narrow range, but the approximation should be recognized.
Step-by-step reasoning
1. Identify initial and final phases with temperature and pressure. 2. Select the correct phase-change direction and molar enthalpy. 3. Reverse sign if the process is freezing, condensation or deposition. 4. Multiply by moles undergoing the change. 5. Add sensible-heating terms and match temperatures for Hess cycles.
Visual explanation
Draw a triangle with solid, liquid and gas at its corners. Label solid→liquid fusion, liquid→gas vaporization and solid→gas sublimation. Add a plus sign showing ΔH sub = ΔH fus + ΔH vap for matched states, and reverse arrows with negative signs.
Real-world analogy
Moving from ground floor to third floor can be done directly or through the second floor; total elevation gain is the same for matching endpoints. Phase-change enthalpies add in an analogous state-function cycle, though heating steps may be needed if transition temperatures differ.
Real-world example
Sweat evaporating from skin absorbs energy, producing cooling. The energy is associated with liquid water becoming vapor, not with breaking water molecules into hydrogen and oxygen. The cooling rate depends on airflow and humidity as well as thermodynamics.
Why?
Why does temperature remain nearly constant while a pure substance melts at fixed pressure? Added energy changes phase proportions and intermolecular organization rather than raising the temperature until melting is complete.
Common misconception
“Vaporization breaks O–H covalent bonds in water.” Water molecules remain H₂O; the major change is separation of molecules and associated interactions, with gas-phase volume effects.
Worked example
Suppose at one chosen common reference temperature a hypothetical substance has ΔH fus = +8 kJ mol⁻¹ and ΔH vap = +30 kJ mol⁻¹. Hess's law gives ΔH sub = +38 kJ mol⁻¹ for solid → gas between matching states. Condensation gas → liquid has ΔH = −30 kJ mol⁻¹. For 0.50 mol sublimed, the enthalpy change is +19 kJ under the same conditions.
Quick check
1. What sign does condensation enthalpy have relative to vaporization? Answer: Equal magnitude and opposite sign at the same stated conditions, so condensation is negative if vaporization is positive.
Exam focus
Keep phase labels and transition conditions explicit. Use latent-heat terms when temperature remains constant during a phase change; do not replace them with mcΔT.
Advanced insight
At equilibrium, the Gibbs energies of coexisting phases are equal, while enthalpy and entropy can differ. For a reversible phase transition at temperature T, ΔS transition = ΔH transition/T. This connects latent heat to the entropy gained on melting or vaporization.
Summary
Fusion, vaporization and sublimation change enthalpy without changing chemical formula. Reverse transitions reverse signs, and Hess's law adds matched phase paths. Heating across a transition requires both temperature-change and latent-heat terms.
Practice questions
1. If ΔH vap = +40 kJ mol⁻¹, what is ΔH for condensing 2 mol at the same conditions? Answer: 2(−40) = −80 kJ. 2. Why can q be nonzero while ΔT is zero during melting? Answer: Energy changes the phase rather than raising temperature during the equilibrium transition. 3. What extra information is needed to add tabulated fusion and vaporization values measured at different temperatures accurately? Answer: Heat capacities or other temperature corrections to express both stages between matched states.