Energy of Phase Changes: Study Guide (Chemistry 6.5)

What happens to energy in a phase change

In Topic 6.4 you used q = mcΔT for a substance that warms up or cools down without changing phase. This topic deals with the second way a system can change its energy: by changing phase.

Picture ice at 0 °C on a warm plate. Energy flows from the plate into the ice. The water molecules in the solid are held in fixed positions by hydrogen bonds. To melt, some of those attractions must be overcome so the molecules can slide past each other. That takes energy.

• Melting (solid → liquid), vaporization (liquid → gas) and sublimation (solid → gas) need energy. The energy of the system increases. They are endothermic.
• Freezing (liquid → solid), condensation (gas → liquid) and deposition (gas → solid) release energy as attractions form. The energy of the system decreases. They are exothermic.

Why the temperature stays constant

Temperature measures the average kinetic energy of the particles. While ice melts at 0 °C, the energy you supply goes into pulling molecules away from their neighbours, which raises their potential energy. Their average speed does not change, so the temperature does not change.

So, for a pure substance at constant pressure:

During a phase change, the temperature stays constant until the change is complete, even though energy is still flowing in (or out).

Only when all of the ice has melted does the added energy start to make the molecules move faster again, and the temperature starts to rise.

Molar enthalpy of a phase change

The energy needed depends on how much substance changes phase. Chemists quote it per mole, at constant pressure, as a molar enthalpy with units of kJ mol⁻¹:

• Molar enthalpy of fusion, ΔH_fus: energy absorbed when 1 mol of solid melts.
• Molar enthalpy of vaporization, ΔH_vap: energy absorbed when 1 mol of liquid vaporizes.

Both are positive because both processes are endothermic. The energy for any amount is then:

q = n × ΔH

where q is in kJ, n is the amount in moles, and ΔH is the molar enthalpy in kJ mol⁻¹. Notice that the mass enters only through n = m / M. There is no ΔT in this equation, because the temperature does not change.

Substance	Phase change	Temperature	Molar enthalpy
Water	Fusion	0 °C	ΔH_fus = +6.01 kJ mol⁻¹
Water	Vaporization	100 °C	ΔH_vap = +40.7 kJ mol⁻¹
Ethanol	Vaporization	78 °C	ΔH_vap = +38.6 kJ mol⁻¹

Opposite changes, opposite signs

Energy is conserved. If 1 mol of water absorbs 40.7 kJ to become steam, then 1 mol of steam must release exactly 40.7 kJ when it condenses back to liquid at the same temperature. So:

• ΔH_condensation = −ΔH_vap. For water at 100 °C, ΔH_cond = −40.7 kJ mol⁻¹.
• ΔH_freezing = −ΔH_fus. For water at 0 °C, ΔH_freezing = −6.01 kJ mol⁻¹.

Data tables usually list only ΔH_fus and ΔH_vap. You use the same numbers for freezing and condensing and change the sign. A negative q means the system released that energy to the surroundings.

Why ΔH_vap is much larger than ΔH_fus

For water, ΔH_vap is about 6.8 times ΔH_fus. Melting only loosens the arrangement: in the liquid the molecules are still touching and still attract each other strongly. Vaporizing separates the molecules completely, so almost all of the remaining attractions must be overcome. This is true for nearly every substance.

The size of ΔH_vap is also a measure of the strength of intermolecular forces (Topic 3.1). Stronger attractions between particles mean more energy is needed per mole to pull them apart, so ΔH_vap is larger.

Reading a heating curve

A heating curve plots the temperature of a sample against the energy added to it (or against time, if energy is added at a steady rate). It combines both kinds of energy change in one picture.

How to read it:

• Sloping sections (A, C, E): one phase only. Energy raises the average kinetic energy, so the temperature rises. Use q = mcΔT here.
• Flat sections (B, D): two phases are present together. Energy changes the potential energy of the particles, so the temperature is constant. Use q = n × ΔH here.
• The temperature of a flat section is the melting point (B) or the boiling point (D) at that pressure.
• The length of a flat section along the energy axis is the energy for that phase change. Section D is much longer than B because ΔH_vap is larger than ΔH_fus.
• A steeper sloping section means less energy is needed per degree, so that phase has a smaller heat capacity. In Figure 1, the liquid section C is the least steep.

A cooling curve is the same picture run backwards: the sample loses energy, and the flat sections are now condensing and freezing.

Worked example 1: boiling a known mass

Question. How much energy is needed to vaporize 25.0 g of ethanol, C₂H₅OH, at its boiling point (78 °C)? How much energy is released when the vapour condenses again at 78 °C? ΔH_vap(ethanol) = 38.6 kJ mol⁻¹.

1. Molar mass: M = 2(12.01) + 6(1.008) + 16.00 = 46.068 g mol⁻¹.
2. Moles: n = 25.0 g ÷ 46.068 g mol⁻¹ = 0.5427 mol.
3. Energy to vaporize: q = n × ΔH_vap = 0.5427 mol × 38.6 kJ mol⁻¹ = 20.95 kJ.
4. Condensing: ΔH_cond = −38.6 kJ mol⁻¹, so q = 0.5427 mol × (−38.6 kJ mol⁻¹) = −20.95 kJ.

Answer. Vaporizing absorbs +20.9 kJ; condensing releases the same amount, q = −20.9 kJ (3 significant figures).

Check. 25 g is about half a mole, and half of 38.6 is about 19, so roughly 20 kJ is sensible. There is no temperature change in this question, so q = mcΔT is not used.

Worked example 2: a calculation with several steps

Question. 30.0 g of ice at −12.0 °C is warmed until it is liquid water at 40.0 °C. Calculate the total energy needed. Use c(ice) = 2.09 J g⁻¹ °C⁻¹, c(water) = 4.18 J g⁻¹ °C⁻¹ and ΔH_fus = 6.01 kJ mol⁻¹.

Break the heating curve into its sections. The ice must first reach 0 °C before it can melt.

Step	What happens	Equation	Working	Energy
1	Ice warms from −12.0 °C to 0.0 °C	q = mcΔT	30.0 × 2.09 × 12.0	752.4 J
2	Ice melts at 0.0 °C	q = n × ΔH_fus	(30.0 ÷ 18.016) mol × 6.01 kJ mol⁻¹ = 1.6652 × 6.01	10.008 kJ
3	Water warms from 0.0 °C to 40.0 °C	q = mcΔT	30.0 × 4.18 × 40.0	5016 J

Add, in the same unit: 0.7524 kJ + 10.008 kJ + 5.016 kJ = 15.776 kJ.

Answer. 15.8 kJ of energy must be absorbed.

Point to notice. The melting step alone is about 63% of the total, even though the temperature does not change during it. Leaving out a phase change is the most common way to lose most of the marks in a question like this. Also notice that steps 1 and 3 use different specific heat capacities: ice and liquid water are different phases.

Worked example 3: finding molar enthalpies from a heating curve

Question. Figure 1 shows the heating curve for 0.500 mol of a fictional substance Q. (a) Find the melting point and boiling point of Q. (b) Calculate ΔH_fus and ΔH_vap for Q. (c) What is the energy change when 0.500 mol of gaseous Q condenses at its boiling point?

(a) The flat sections are at −10 °C (melting point) and 50 °C (boiling point).

(b) Read the length of each flat section along the energy axis.

1. Melting (B) runs from 1.2 kJ to 4.8 kJ: q = 3.6 kJ for 0.500 mol. ΔH_fus = q ÷ n = 3.6 kJ ÷ 0.500 mol = 7.2 kJ mol⁻¹.
2. Boiling (D) runs from 10.8 kJ to 25.8 kJ: q = 15.0 kJ for 0.500 mol. ΔH_vap = 15.0 kJ ÷ 0.500 mol = 30.0 kJ mol⁻¹.

(c) Condensing is the reverse of boiling, so q = −15.0 kJ (ΔH_cond = −30.0 kJ mol⁻¹). The system releases 15.0 kJ to its surroundings.

Check. ΔH_vap is about four times ΔH_fus, which fits the pattern that vaporizing separates particles completely while melting only loosens them. Remember to divide by the moles in the sample: the length of the flat section is the energy for 0.500 mol, not for 1 mol.

Common misconceptions

• “The temperature keeps rising while water boils, as long as you keep heating.” For a pure substance at constant pressure the temperature stays at the boiling point until all the liquid has gone. Extra heating makes it boil faster, not hotter.
• “During melting the particles speed up.” Their average kinetic energy is constant. The energy raises their potential energy as attractions are overcome.
• “Freezing needs energy because you put water in a freezer.” The freezer removes energy from the water. Freezing is exothermic for the water: the water releases energy to the colder surroundings.
• Using mass instead of moles with ΔH. ΔH_fus and ΔH_vap are per mole. Multiplying 18 g of water by 6.01 kJ mol⁻¹ gives a number 18 times too large.
• Mixing J and kJ. q = mcΔT gives joules with the usual heat capacity units; q = n × ΔH gives kilojoules. Convert before you add.
• Using q = mcΔT across a phase change. The equation applies to one phase over a temperature range. Split the problem at every melting or boiling point.
• Forgetting the sign when reversing. Condensing and freezing are exothermic, so their q values are negative.
• “Boiling breaks covalent bonds.” When water boils, molecules of H₂O separate from each other. The O–H bonds inside each molecule are not broken.

Where this leads

Phase changes are one of three ways a system changes its energy; heating and cooling (Topic 6.4) is another. The third is a chemical reaction, where q depends on the moles reacting and the molar enthalpy of reaction. That is Topic 6.6, Introduction to Enthalpy of Reaction. Try the practice questions now, then use the revision notes and the checklist to consolidate.