DAT General Chemistry: Enthalpy and Entropy of Phase Changes

DAT High-Yield Concept

Endothermic Phase Changes: Fusion, vaporization, and sublimation break or disrupt intermolecular forces; they require energy input ( $\Delta H > 0$ ) and increase molecular disorder ( $\Delta S > 0$ ).
Exothermic Phase Changes: Freezing, condensation, and deposition form intermolecular interactions; they release heat ( $\Delta H < 0$ ) and decrease entropy ( $\Delta S < 0$ ).
Magnitude Relationship: Enthalpy of vaporization ( $\Delta H_{\text{vap}}$ ) is substantially larger than enthalpy of fusion ( $\Delta H_{\text{fus}}$ ) because vaporization requires overcoming intermolecular attractions entirely.
Phase Equilibrium: During an isobaric, isothermal phase coexistence, $\Delta G = 0$ , yielding the thermodynamic expression $\Delta S_{\text{trans}} = \frac{\Delta H_{\text{trans}}}{T_{\text{trans}}}$ .

Thermodynamic Signs of Phase Transitions

Phase transitions are physical transformations between solid, liquid, and gaseous states driven by changes in thermal energy and entropy. When a substance moves to a phase with greater molecular mobility (solid to liquid to gas), energy must be absorbed to overcome intermolecular forces. Conversely, transitions toward a more ordered state (gas to liquid to solid) release energy as intermolecular contacts are established.

Phase Transition	Representative Reaction	Enthalpy Sign ( $\Delta H$ )	Entropy Sign ( $\Delta S$ )
Fusion (Melting)	$\ce{H2O(s) -> H2O(l)}$	$\Delta H_{\text{fus}} > 0$ (Endothermic)	$\Delta S_{\text{fus}} > 0$ (Increasing disorder)
Freezing (Solidification)	$\ce{H2O(l) -> H2O(s)}$	$\Delta H_{\text{freeze}} < 0$ (Exothermic)	$\Delta S_{\text{freeze}} < 0$ (Decreasing disorder)
Vaporization	$\ce{H2O(l) -> H2O(g)}$	$\Delta H_{\text{vap}} > 0$ (Endothermic)	$\Delta S_{\text{vap}} > 0$ (Increasing disorder)
Condensation	$\ce{H2O(g) -> H2O(l)}$	$\Delta H_{\text{cond}} < 0$ (Exothermic)	$\Delta S_{\text{cond}} < 0$ (Decreasing disorder)
Sublimation	$\ce{H2O(s) -> H2O(g)}$	$\Delta H_{\text{sub}} > 0$ (Endothermic)	$\Delta S_{\text{sub}} > 0$ (Increasing disorder)
Deposition	$\ce{H2O(g) -> H2O(s)}$	$\Delta H_{\text{dep}} < 0$ (Exothermic)	$\Delta S_{\text{dep}} < 0$ (Decreasing disorder)

Memory Aid: "Moving UP the state stairs (Solid → Liquid → Gas) is ENDO (+ $\Delta H$ , + $\Delta S$ ); stepping DOWN is EXO (- $\Delta H$ , - $\Delta S$ )."

Thermodynamic Equilibrium at Transition Temperature

At constant pressure and at the exact transition temperature ( $T_{\text{trans}}$ ), two coexisting phases are in dynamic equilibrium, meaning $\Delta G = 0$ . Because $\Delta G = \Delta H - T\Delta S = 0$ , we derive:

\Delta S_{\text{trans}} = \frac{\Delta H_{\text{trans}}}{T_{\text{trans}}}

Note: $T_{\text{trans}}$ must always be expressed in Kelvin ( $\pu{K}$ ).

Enthalpy of Fusion and Vaporization

The molar enthalpy of fusion ( $\Delta H_{\text{fus}}$ ) is the heat required to convert one mole of solid to liquid at constant pressure. The molar enthalpy of vaporization ( $\Delta H_{\text{vap}}$ ) is the heat required to convert one mole of liquid to gas.

For any substance, $\Delta H_{\text{vap}}$ is significantly greater than $\Delta H_{\text{fus}}$ . Melting only partially disrupts the crystalline lattice to allow fluid motion, whereas vaporization requires completely separating molecules against attractive intermolecular forces into the gas phase.

Worked Calculation 1: Heat Required for Vaporization

Problem: Calculate the thermal energy in kilojoules required to completely vaporize $\pu{36.0 g}$ of liquid water at $\pu{100 degC}$ . (Given: $\Delta H_{\text{vap}} = \pu{40.7 kJ/mol}$ for $\ce{H2O}$ , molar mass = $\pu{18.0 g/mol}$ ).

Reasoning & Steps:

Convert mass to moles using dimensional analysis: $n = \pu{36.0 g} \times \frac{\pu{1 mol}}{\pu{18.0 g}} = \pu{2.00 mol}$
Calculate heat absorbed ( $q$ ) during the isothermal phase transition: $q = n \cdot \Delta H_{\text{vap}} = \pu{2.00 mol} \times \pu{40.7 kJ/mol} = \pu{81.4 kJ}$

Common Exam Trap: Do not use $q = m c \Delta T$ during a phase change. Phase transitions occur isothermally (constant temperature), so $\Delta T = 0$ ; instead, use $q = n \Delta H$ or $q = m \Delta H$ .

Worked Calculation 2: Molar Entropy of Fusion

Problem: Determine the molar entropy of fusion ( $\Delta S_{\text{fus}}$ ) for water at its normal melting point of $\pu{0 degC}$ ( $\pu{273.15 K}$ ), given that $\Delta H_{\text{fus}} = \pu{6.01 kJ/mol}$ .

Reasoning & Steps:

Convert enthalpy from $\pu{kJ/mol}$ to $\pu{J/mol}$ : $\Delta H_{\text{fus}} = \pu{6.01 kJ/mol} \times \frac{\pu{1000 J}}{\pu{1 kJ}} = \pu{6010 J/mol}$
Apply the equilibrium relation $\Delta S_{\text{fus}} = \frac{\Delta H_{\text{fus}}}{T_{\text{fus}}}$ : $\Delta S_{\text{fus}} = \frac{\pu{6010 J/mol}}{\pu{273.15 K}} = \pu{22.0 J/(mol K)}$

Common Exam Trap: Forgetting to convert temperature to Kelvin yields a division by zero at $\pu{0 degC}$ . Always convert Celsius to Kelvin ( $T_{\text{K}} = T_{\text{C}} + 273.15$ ).

DAT Critical Thinking Questions

Why does steam at $\pu{100 degC}$ cause more severe thermal burns to human skin than liquid water at $\pu{100 degC}$ ?
How would you expect the magnitude of $\Delta H_{\text{vap}}$ for ethanol ( $\ce{C2H5OH}$ ) to compare to that of dimethyl ether ( $\ce{CH3OCH3}$ ), given their structural isomers?
Under isobaric conditions below a substance's triple point, why is sublimation spontaneous at elevated temperatures despite having a positive enthalpy change ( $\Delta H_{\text{sub}} > 0$ )?

References

Sources returned by Google Search grounding during lesson verification.

11.4: Phase Changes - Chemistry LibreTexts — chem.libretexts.org
Enthalpy and phase changes - Khan Academy — khanacademy.org

Enthalpy and Entropy of Phase Changes

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