Chemistry·Revision Notes

Enthalpy of Phase Transition — Revision Notes

NEET UG
Version 1Updated 22 Mar 2026

⚡ 30-Second Revision

Key Formulas & Concepts:

  • Phase Change:Constant T, energy for IMFs.
  • Heating/Cooling:q=mcdotccdotDeltaTq = m cdot c cdot Delta T (temperature change).
  • Fusion (Melting):Solid ightarrowightarrow Liquid, DeltaHfus>0Delta H_{fus} > 0. q=ncdotDeltaHfusq = n cdot Delta H_{fus} (or mcdotLfusm cdot L_{fus}).
  • Vaporization (Boiling):Liquid ightarrowightarrow Gas, DeltaHvap>0Delta H_{vap} > 0. q=ncdotDeltaHvapq = n cdot Delta H_{vap} (or mcdotLvapm cdot L_{vap}).
  • Sublimation:Solid ightarrowightarrow Gas, DeltaHsub>0Delta H_{sub} > 0. DeltaHsub=DeltaHfus+DeltaHvapDelta H_{sub} = Delta H_{fus} + Delta H_{vap}.
  • Reverse Processes:Freezing, Condensation, Deposition are exothermic (DeltaH<0Delta H < 0).

* Freezing: DeltaHfus-Delta H_{fus} * Condensation: DeltaHvap-Delta H_{vap} * Deposition: DeltaHsub-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, DeltaHfusDelta H_{fus}), vaporization (boiling, liquid to gas, DeltaHvapDelta H_{vap}), and sublimation (solid to gas, DeltaHsubDelta H_{sub}).

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

For NEET, expect numerical problems combining specific heat calculations (q=mcDeltaTq = mcDelta T) for temperature changes within a phase, and latent heat calculations (q=mLq = mL or q=nDeltaHtransitionq = 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 0circC0^circ\text{C} requires the enthalpy of fusion (DeltaHfusDelta H_{fus}), an endothermic process where energy breaks the solid lattice. Boiling water at 100circC100^circ\text{C} requires the enthalpy of vaporization (DeltaHvapDelta H_{vap}), an even more endothermic process to fully separate liquid molecules into gas.

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

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

Worked Mini-Example: How much heat is needed to convert 1,mol1,\text{mol} of water from liquid at 25circC25^circ\text{C} to steam at 100circC100^circ\text{C}? (Given: cwater=75.3,J/molcircCc_{water} = 75.3,\text{J/mol}^circ\text{C}, DeltaHvap=40.7,kJ/molDelta H_{vap} = 40.7,\text{kJ/mol})

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  1. **Heat water from 25circC25^circ\text{C} to 100circC100^circ\text{C}:**

q1=ncdotcwatercdotDeltaT=1,mol×75.3,J/molcircC×(10025)circC=5647.5,J=5.6475,kJq_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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  1. **Vaporize water at 100circC100^circ\text{C}:**

q2=ncdotDeltaHvap=1,mol×40.7,kJ/mol=40.7,kJq_2 = n cdot Delta H_{vap} = 1,\text{mol} \times 40.7,\text{kJ/mol} = 40.7,\text{kJ}

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  1. Total Heat:qtotal=q1+q2=5.6475,kJ+40.7,kJ=46.3475,kJq_{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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  1. Definition:Enthalpy of phase transition (DeltaHtransitionDelta 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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  3. Purpose of Energy:Energy is used to overcome/establish intermolecular forces, NOT to change kinetic energy (temperature).
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  5. **Endothermic Processes (DeltaH>0Delta H > 0, heat absorbed):**

* Fusion (Melting): Solid ightarrowightarrow Liquid (DeltaHfusDelta H_{fus}) * Vaporization (Boiling): Liquid ightarrowightarrow Gas (DeltaHvapDelta H_{vap}) * Sublimation: Solid ightarrowightarrow Gas (DeltaHsubDelta H_{sub})

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  1. **Exothermic Processes (DeltaH<0Delta H < 0, heat released):**

* Freezing: Liquid ightarrowightarrow Solid (DeltaHfus-Delta H_{fus}) * Condensation: Gas ightarrowightarrow Liquid (DeltaHvap-Delta H_{vap}) * Deposition: Gas ightarrowightarrow Solid (DeltaHsub-Delta H_{sub})

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  1. Hess's Law Relation:DeltaHsub=DeltaHfus+DeltaHvapDelta H_{sub} = Delta H_{fus} + Delta H_{vap}.
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  3. Formulas for Heat Calculation:

* For temperature change (within a phase): q=mcdotccdotDeltaTq = m cdot c cdot Delta T (where mm is mass, cc is specific heat capacity, DeltaTDelta T is temperature change). * For phase change (at constant temperature): q=ncdotDeltaHtransitionq = n cdot Delta H_{transition} (where nn is moles, DeltaHtransitionDelta H_{transition} is molar enthalpy of transition) OR q=mcdotLtransitionq = m cdot L_{transition} (where LtransitionL_{transition} is latent heat per gram).

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  1. Factors Affecting Magnitude:Stronger intermolecular forces lead to higher DeltaHfusDelta H_{fus} and DeltaHvapDelta H_{vap} values.
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  3. 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, DeltaH>0Delta H > 0). Their opposites (Freezing, Condensation, Deposition) release energy (exothermic, DeltaH<0Delta H < 0).

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

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