AP Chemistry 6.4: Energy of Phase Changes

Changing the physical state of matter requires energy because particle attractions must be disrupted or allowed to form.

• Khan Academy's current lesson specifically covers enthalpy of phase changes and heating curves.

The following phase changes require energy:

• melting: solid → liquid
• vaporization: liquid → gas
• sublimation: solid → gas
• During these processes, particles become less strongly constrained by intermolecular attractions.
• Therefore:
• ΔH > 0

The reverse phase changes release energy:

• freezing: liquid → solid
• condensation: gas → liquid
• deposition: gas → solid
• Therefore:
• ΔH < 0

The energy required for a phase change has the same magnitude as the energy released by the reverse phase change.

• For example, if:
• ΔHvap = +40.7 kJ/mol
• then:
• ΔHcond = −40.7 kJ/mol
• The sign changes because the direction of energy transfer reverses.

For many substances:

• ΔHvap > ΔHfus

During melting, intermolecular attractions are weakened enough for particles to move past one another, but the particles still remain relatively close.

During vaporization, particles must become widely separated from one another, meaning much more of the intermolecular attraction must be overcome.

• This usually requires more energy.

If a molar enthalpy of phase change is given:

• q = nΔHphase
• where:
• n = moles
• ΔHphase = molar enthalpy of fusion, vaporization, etc.
• Suppose 2.00 mol of a liquid vaporizes and:
• ΔHvap = +35.0 kJ/mol
• Then:
• q = (2.00 mol)(35.0 kJ/mol)
• q = +70.0 kJ
• The positive sign makes sense because vaporization is endothermic.

When a pure substance changes phase at constant pressure, temperature remains approximately constant during the phase transition.

• This can initially seem strange because energy is still being added.
• The added energy is being used primarily to change potential energy associated with particle attractions, rather than increasing average kinetic energy.
• Since temperature reflects average kinetic energy, the temperature remains constant during the phase change.

A heating curve shows how the temperature of a substance changes as energy is continuously added.

• A typical heating curve contains sloped sections and flat sections.

During a sloped portion:

• temperature changes
• but:
• phase stays the same
• For these portions:
• q = mcΔT
• Examples include:
• heating solid
• heating liquid
• heating gas
• During these sections, average kinetic energy increases.

During a plateau:

• temperature stays constant
• while:
• phase changes
• For these portions:
• q = nΔHphase

During melting:

• solid + liquid coexist.

During boiling:

• liquid + gas coexist.
• The energy added during a plateau increases potential energy by overcoming intermolecular attractions.

A problem might ask how much energy is needed to heat ice below 0°C until it becomes steam above 100°C.

• You cannot use one q=mcΔT calculation for the entire process.
• Instead, divide it into steps.
• Example path:

Step 1: heat solid

• q₁ = mcsolidΔT

Step 2: melt solid

• q₂ = nΔHfus

Step 3: heat liquid

• q₃ = mcliquidΔT

Step 4: vaporize liquid

• q₄ = nΔHvap

Step 5: heat gas

• q₅ = mcgasΔT
• Then:
• qtotal = q₁ + q₂ + q₃ + q₄ + q₅

Use only the steps actually present in the problem.

During sloped regions:

• temperature ↑ → average kinetic energy ↑

During flat regions:

• temperature constant → average kinetic energy approximately constant
• but:
• potential energy changes
• This kinetic-energy vs. potential-energy distinction is frequently useful in AP explanations.

Do not use q=mcΔT during a constant-temperature phase change.

Do not use q=nΔHphase while a substance is simply warming within one phase.

Do not assume temperature rises during melting or boiling.

Remember to include every relevant segment in a multi-step heating problem.

On a heating curve:

• slope = temperature change → q = mcΔT
• plateau = phase change → q = nΔHphase

Opposite processes have enthalpy changes with equal magnitudes and opposite signs.

• For example:
• ΔHvap = −ΔHcond
• and:
• ΔHfus = −ΔHfreezing