Interconversion of States

When a solid absorbs sufficient heat energy, its particles gain enough kinetic energy to overcome the rigid lattice forces holding them in fixed positions. The temperature at which this occurs is the melting point (Tm). For ice, Tm = 0 °C at 1 atm. The energy required per mole is the enthalpy of fusion (ΔHfus); for water, ΔHfus = +6.01 kJ mol⁻¹. Because the energy goes into breaking intermolecular bonds rather than raising temperature, the temperature stays constant during melting.

Freezing is the exact reverse of melting. As a liquid loses heat, particle kinetic energy decreases until intermolecular forces can lock particles into an ordered crystal lattice. The freezing point equals the melting point for a pure substance. The enthalpy of freezing = −ΔHfus (exothermic). This is why a glass of ice water stays at 0 °C until all the ice melts — energy is continuously being released as liquid water freezes onto the ice surface.

Vaporisation occurs at the surface of a liquid at any temperature (evaporation); boiling occurs throughout the liquid when the vapour pressure equals the external pressure. The boiling point (Tb) is therefore pressure-dependent: water boils at 100 °C at 1 atm but only ~70 °C at 0.31 atm (high altitude). The enthalpy of vaporisation (ΔHvap) for water is +40.7 kJ mol⁻¹ — much larger than ΔHfus because all remaining intermolecular hydrogen bonds must be broken. This large value makes water an excellent coolant.

Condensation is the reverse of vaporisation. Gas molecules lose kinetic energy, slow down, and are captured by intermolecular attractive forces to form liquid droplets. The enthalpy of condensation = −ΔHvap (exothermic). Atmospheric condensation of water vapour forms clouds and rain. In the laboratory, a condenser cools vapour with cold water flowing around the outside, converting distillate vapour back to liquid for collection.

Sublimation occurs when the solid-phase vapour pressure is high enough that particles escape directly to the gas phase without passing through the liquid state. This happens below the triple point of a substance. Dry ice (solid CO₂) sublimes at −78.5 °C at 1 atm because the triple point of CO₂ is at 5.18 atm — CO₂ liquid cannot exist at atmospheric pressure. Iodine (I₂) also sublimes readily, producing a distinctive purple vapour. The enthalpy of sublimation = ΔHfus + ΔHvap (Hess's law).

Deposition is the reverse of sublimation: gas particles lose enough energy to form an ordered solid directly. Frost on a cold window forms by deposition of water vapour, not by freezing of liquid water. Snow crystals also form by deposition of water vapour onto dust particles in clouds. This process is exploited in physical vapour deposition (PVD), an industrial technique for coating surfaces with thin films of metals or ceramics.

A phase diagram maps which state is stable at any given temperature (T) and pressure (P). The lines separating regions are equilibrium curves where two phases coexist. The triple point is the unique T-P combination where all three phases coexist (water: 0.01 °C, 611.7 Pa). The critical point marks the end of the liquid–gas boundary; above it, the substance becomes a supercritical fluid with properties of both states. Water has an unusual negative slope on its solid–liquid boundary, meaning increasing pressure lowers the melting point — a consequence of ice being less dense than liquid water.

• q — Heat energy transferred (J or kJ): Positive (endothermic) for melting/vaporisation/sublimation; negative (exothermic) for freezing/condensation/deposition.
• n — Amount of substance (mol): The number of moles of the pure substance undergoing the phase change.
• \Delta H_{\text{fus}} — Molar enthalpy of fusion (kJ mol⁻¹): Energy required to melt one mole of substance at its melting point; e.g., +6.01 kJ mol⁻¹ for water.
• \Delta H_{\text{vap}} — Molar enthalpy of vaporisation (kJ mol⁻¹): Energy required to vaporise one mole of substance at its boiling point; e.g., +40.7 kJ mol⁻¹ for water.

Problem. How much heat energy (in kJ) is required to completely melt 36.0 g of ice at 0 °C? (ΔHfus of water = 6.01 kJ mol⁻¹; molar mass of H₂O = 18.0 g mol⁻¹)

Answer. 12.0 kJ of heat energy must be absorbed to melt 36.0 g of ice at 0 °C.

• Kinetic Molecular Theory — KMT explains why particles in different states have different energies and why temperature drives phase transitions by changing average kinetic energy.
• Intermolecular Forces — The strength of intermolecular forces (hydrogen bonding, van der Waals, dipole–dipole) determines the melting and boiling points of substances and the energy needed for phase transitions.
• Enthalpy and Calorimetry — Latent heats (ΔHfus, ΔHvap) are measured using calorimetry and are central to calculating heat changes during phase transitions.
• Vapour Pressure and Boiling Point — Understanding vapour pressure is essential for explaining why boiling point depends on external pressure and how the Clausius–Clapeyron equation links pressure and temperature at phase boundaries.