Enthalpy of Phase Transition — Revision Notes

⚡ 30-Second Revision

Key Formulas & Concepts:

Phase Change: — Constant T, energy for IMFs.
Heating/Cooling: — $q = m \cdot c \cdot \Delta T$ (temperature change).
Fusion (Melting): — Solid $\rightarrow$ Liquid, $\Delta H_{fus} > 0$. $q = n \cdot \Delta H_{fus}$ (or $m \cdot L_{fus}$).
Vaporization (Boiling): — Liquid $\rightarrow$ Gas, $\Delta H_{vap} > 0$. $q = n \cdot \Delta H_{vap}$ (or $m \cdot L_{vap}$).
Sublimation: — Solid $\rightarrow$ Gas, $\Delta H_{sub} > 0$. $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$.
Reverse Processes: — Freezing, Condensation, Deposition are exothermic ($\Delta H < 0$).

* Freezing: $-\Delta H_{fus}$ * Condensation: $-\Delta H_{vap}$ * Deposition: $-\Delta H_{sub}$

Units: — Be careful with J vs. kJ, g vs. mol.

2-Minute Revision

Enthalpy of phase transition refers to the heat absorbed or released when a substance changes its physical state at constant temperature and pressure. This 'latent heat' is used to overcome or form intermolecular forces, not to change the kinetic energy of particles.

Key endothermic transitions (requiring heat) are fusion (melting, solid to liquid, $\Delta H_{fus}$), vaporization (boiling, liquid to gas, $\Delta H_{vap}$), and sublimation (solid to gas, $\Delta H_{sub}$).

Their reverse processes – freezing, condensation, and deposition – are exothermic (releasing heat) and have negative enthalpy changes. The relationship $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$ is crucial.

For NEET, expect numerical problems combining specific heat calculations ($q = mcDelta T$) for temperature changes within a phase, and latent heat calculations ($q = mL$ or $q = nDelta H_{transition}$) for phase changes.

Always pay attention to units (J/g vs. kJ/mol) and ensure all steps are accounted for in multi-stage problems.

5-Minute Revision

To master enthalpy of phase transition for NEET, focus on the underlying principles and problem-solving methodology. A phase transition is a physical change where a substance alters its state of matter (solid, liquid, gas).

Crucially, these occur at constant temperature (e.g., melting point, boiling point) because the energy exchanged (latent heat) is entirely dedicated to overcoming or establishing intermolecular forces, not increasing particle kinetic energy.

For instance, melting ice at $0^\circ\text{C}$ requires the enthalpy of fusion ($\Delta H_{fus}$), an endothermic process where energy breaks the solid lattice. Boiling water at $100^\circ\text{C}$ requires the enthalpy of vaporization ($\Delta H_{vap}$), an even more endothermic process to fully separate liquid molecules into gas.

Sublimation ($\Delta H_{sub}$) is the direct solid-to-gas transition, related by Hess's Law: $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$.

Remember that reverse processes are exothermic: freezing (liquid to solid) releases $-\Delta H_{fus}$, condensation (gas to liquid) releases $-\Delta H_{vap}$, and deposition (gas to solid) releases $-\Delta H_{sub}$. The magnitude of these enthalpy changes is directly proportional to the strength of intermolecular forces. Stronger IMFs mean higher $\Delta H_{fus}$ and $\Delta H_{vap}$.

Worked Mini-Example: How much heat is needed to convert $1\,\text{mol}$ of water from liquid at $25^\circ\text{C}$ to steam at $100^\circ\text{C}$? (Given: $c_{water} = 75.3\,\text{J/mol}^\circ\text{C}$, $\Delta H_{vap} = 40.7\,\text{kJ/mol}$)

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**Heat water from $25^\circ\text{C}$ to $100^\circ\text{C}$:**

$q_1 = n \cdot c_{water} \cdot \Delta T = 1\,\text{mol} \times 75.3\,\text{J/mol}^\circ\text{C} \times (100 - 25)^\circ\text{C} = 5647.5\,\text{J} = 5.6475\,\text{kJ}$

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**Vaporize water at $100^\circ\text{C}$:**

$q_2 = n \cdot \Delta H_{vap} = 1\,\text{mol} \times 40.7\,\text{kJ/mol} = 40.7\,\text{kJ}$

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Total Heat: — $q_{total} = q_1 + q_2 = 5.6475\,\text{kJ} + 40.7\,\text{kJ} = 46.3475\,\text{kJ}$.

Always break down complex problems into these distinct steps, paying close attention to units and the correct application of formulas.

Prelims Revision Notes

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Definition: — Enthalpy of phase transition ($\Delta H_{transition}$) is the heat change at constant T and P during a phase change. It's also called latent heat.
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Purpose of Energy: — Energy is used to overcome/establish intermolecular forces, NOT to change kinetic energy (temperature).
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**Endothermic Processes ($\Delta H > 0$, heat absorbed):**

* Fusion (Melting): Solid $\rightarrow$ Liquid ($\Delta H_{fus}$) * Vaporization (Boiling): Liquid $\rightarrow$ Gas ($\Delta H_{vap}$) * Sublimation: Solid $\rightarrow$ Gas ($\Delta H_{sub}$)

1
**Exothermic Processes ($\Delta H < 0$, heat released):**

* Freezing: Liquid $\rightarrow$ Solid ($-\Delta H_{fus}$) * Condensation: Gas $\rightarrow$ Liquid ($-\Delta H_{vap}$) * Deposition: Gas $\rightarrow$ Solid ($-\Delta H_{sub}$)

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Hess's Law Relation: — $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$.
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Formulas for Heat Calculation:

* For temperature change (within a phase): $q = m \cdot c \cdot \Delta T$ (where $m$ is mass, $c$ is specific heat capacity, $\Delta T$ is temperature change). * For phase change (at constant temperature): $q = n \cdot \Delta H_{transition}$ (where $n$ is moles, $\Delta H_{transition}$ is molar enthalpy of transition) OR $q = m \cdot L_{transition}$ (where $L_{transition}$ is latent heat per gram).

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Factors Affecting Magnitude: — Stronger intermolecular forces lead to higher $\Delta H_{fus}$ and $\Delta H_{vap}$ values.
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Common Mistakes: — Forgetting a step in multi-stage problems, incorrect unit conversions (J to kJ, g to mol), misidentifying endothermic/exothermic processes, confusing specific heat with latent heat.

Vyyuha Quick Recall

To remember the endothermic phase changes: My Very Solid Substance Melts, Vaporizes, Sublimes.

Melts (Fusion)
Vaporizes (Vaporization)
Sublimes (Sublimation)

All these processes require energy input (endothermic, $\Delta H > 0$). Their opposites (Freezing, Condensation, Deposition) release energy (exothermic, $\Delta H < 0$).

For the relationship: Sublimation Is Fusion Plus Vaporization (SIFPV) -> $\Delta H_{sub} = \Delta H_{fus} + \Delta H_{vap}$.