TutorChase logo
Login
AP Chemistry Notes

6.5.4 Using Molar Enthalpies (Fusion and Vaporization) in Calculations

AP Syllabus focus: ‘Molar enthalpy of fusion and vaporization can be used to calculate energy absorbed during melting/boiling and released during freezing/condensing.’

Phase changes involve large energy transfers without changing temperature.

Pasted image

Simplified phase diagram of water showing the regions where solid, liquid, and vapor are stable, separated by equilibrium boundaries (fusion, vaporization, and sublimation curves). The labeled triple point and critical point provide anchor concepts for when phase boundaries (where ΔHfus\Delta H_{fus} and ΔHvap\Delta H_{vap} apply) are encountered. Source

In AP Chemistry, you quantify these transfers using molar enthalpy of fusion and molar enthalpy of vaporization, scaled by the amount of substance undergoing the phase change.

Molar enthalpies used for phase-change calculations

Molar enthalpy of fusion, ΔHfus

Molar enthalpy of fusion (ΔHfus\Delta H_{fus}): The energy absorbed to melt 1 mole of a substance (solid \rightarrow liquid) at its melting point, at constant pressure.

Key points for using ΔHfus\Delta H_{fus}:

  • Applies only to the solid–liquid phase change.

  • Reported in kJ/mol (or J/mol).

  • The tabulated magnitude corresponds to melting; the reverse direction uses the opposite sign.

Molar enthalpy of vaporization, ΔHvap

Molar enthalpy of vaporization (ΔHvap\Delta H_{vap}): The energy absorbed to vaporise 1 mole of a substance (liquid \rightarrow gas) at its boiling point, at constant pressure.

Key points for using ΔHvap\Delta H_{vap}:

  • Applies only to the liquid–gas phase change.

  • Reported in kJ/mol (or J/mol).

  • The tabulated magnitude corresponds to boiling/vaporising; the reverse direction uses the opposite sign.

Core relationship: scaling by moles

In these problems, the phase-change energy is proportional to the number of moles that change phase.

q=nΔHphase q = n\Delta H_{phase}

q q = heat absorbed by the system (positive) or released by the system (negative), in kJ or J

n n = amount of substance undergoing the phase change, in mol

ΔHphase \Delta H_{phase} = molar enthalpy for the relevant phase change (e.g., ΔHfus\Delta H_{fus} or ΔHvap\Delta H_{vap}), in kJ/mol or J/mol

Use this relationship only for the phase-change step itself; temperature change steps are handled differently and should not be mixed into this equation.

Choosing the correct molar enthalpy and sign

To match the syllabus language (“absorbed during melting/boiling” vs “released during freezing/condensing”), decide the process direction first.

  • Melting (solid \rightarrow liquid): energy is absorbed by the system

    • Use +ΔHfus+\Delta H_{fus} in q=nΔHphaseq=n\Delta H_{phase}

  • Freezing (liquid \rightarrow solid): energy is released by the system

    • Use ΔHfus-\Delta H_{fus}

  • Boiling/vaporising (liquid \rightarrow gas): energy is absorbed by the system

    • Use +ΔHvap+\Delta H_{vap}

  • Condensing (gas \rightarrow liquid): energy is released by the system

    • Use ΔHvap-\Delta H_{vap}

Sign discipline is essential:

  • q>0q>0 means the system absorbs heat (endothermic phase change).

  • q<0q<0 means the system releases heat (exothermic phase change).

Converting the given sample information into moles

The equation requires nn in moles. Most AP questions provide a mass and a molar mass (or enough information to find it).

  • If given mass mm (in g), find moles using:

    • n=mMn = \dfrac{m}{M}

  • Use the moles of the substance actually undergoing the phase change (typically the same chemical formula given for ΔHfus\Delta H_{fus} or ΔHvap\Delta H_{vap}).

Common unit checks:

  • If ΔH\Delta H is in kJ/mol, keep qq in kJ.

  • If ΔH\Delta H is in J/mol, keep qq in J.

  • Convert only once at the end if needed (1 kJ = 1000 J).

What the calculation represents physically

When you compute q=nΔHfusq=n\Delta H_{fus} or q=nΔHvapq=n\Delta H_{vap}, you are accounting for energy associated with overcoming or forming interparticle attractions during a phase change:

  • Melting/boiling: energy goes into separating particles enough to change phase.

  • Freezing/condensing: energy is released as particles come closer and form more stabilising interactions.

These molar enthalpies are used specifically because the temperature does not change during the phase change step; the energy transfer changes the phase, not the kinetic energy distribution.

High-frequency pitfalls to avoid

  • Do not use q=mcΔTq=mc\Delta T for the phase-change segment; use q=nΔHphaseq=n\Delta H_{phase}.

  • Match the correct ΔH\Delta H to the phase boundary:

    • solid–liquid uses ΔHfus\Delta H_{fus}, liquid–gas uses ΔHvap\Delta H_{vap}.

  • Include the correct sign for freezing/condensing (released energy implies negative qq for the system).

  • Use the correct amount: the energy depends on how many moles actually change phase, not the total moles present if the process is partial.

  • Keep significant figures consistent with the given data and avoid mixing kJ and J within the same step.

FAQ

Vaporisation requires much greater separation of particles than melting.

Gas formation typically involves overcoming nearly all intermolecular attractions, whereas melting only partially disrupts them.

$\Delta H_{vap}$ depends on the boiling temperature, which changes with pressure.

At higher external pressure, the boiling point increases; the measured $\Delta H_{vap}$ at that new temperature can differ from the standard tabulated value.

“Per mole” refers only to the amount that actually undergoes the phase change.

If a process stops early, use moles transformed (not initial moles present) when applying $q=n\Delta H_{phase}$.

They are typically obtained by calorimetry at the phase-change temperature.

A known heat input (or output) is related to the measured amount of substance that melts or vaporises at constant pressure.

They are best used near the stated conditions (often 1 bar and the normal melting/boiling point).

Far from those conditions, phase-change enthalpies can vary with temperature and pressure, so additional data or assumptions may be required.

Practice Questions

(1–3 marks) A sample of 12.0 g of water freezes at 00^\circC. Given ΔHfus(H2O)=6.01 kJ mol1\Delta H_{fus}(\text{H}_2\text{O}) = 6.01\ \text{kJ mol}^{-1}, determine the heat change qq for the water (sign included).

  • Convert mass to moles: n=12.018.0=0.667 moln = \dfrac{12.0}{18.0} = 0.667\ \text{mol} (1)

  • Freezing releases heat: use ΔHfus-\Delta H_{fus} (1)

  • q=n(ΔHfus)=0.667×(6.01)=4.01 kJq = n(-\Delta H_{fus}) = 0.667 \times (-6.01) = -4.01\ \text{kJ} (1)

(4–6 marks) A 0.250 mol sample of a substance is first melted, then completely vaporised at its phase-change temperatures. Given ΔHfus=8.5 kJ mol1\Delta H_{fus}=8.5\ \text{kJ mol}^{-1} and ΔHvap=31.2 kJ mol1\Delta H_{vap}=31.2\ \text{kJ mol}^{-1}, calculate the total heat absorbed by the system for both phase changes.

  • Identify both steps are endothermic: qfus=nΔHfusq_{fus}=n\Delta H_{fus} and qvap=nΔHvapq_{vap}=n\Delta H_{vap} (1)

  • Calculate qfus=0.250×8.5=2.13 kJq_{fus}=0.250\times 8.5=2.13\ \text{kJ} (1)

  • Calculate qvap=0.250×31.2=7.80 kJq_{vap}=0.250\times 31.2=7.80\ \text{kJ} (1)

  • Add phase-change heats: qtotal=2.13+7.80=9.93 kJq_{total}=2.13+7.80=9.93\ \text{kJ} (1)

  • State absorbed by system, so positive: +9.93 kJ+9.93\ \text{kJ} (1)

Hire a tutor

Please fill out the form and we'll find a tutor for you.

1/2
Your details
Alternatively contact us via
WhatsApp, Phone Call, or Email