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Phase changes

Covers AS91390

The three enthalpy changes of state

$\Delta_{fus}H^\circ$ — fusion: 1 mol solid → liquid (melting)
$\Delta_{vap}H^\circ$ — vaporisation: 1 mol liquid → gas (boiling)
$\Delta_{sub}H^\circ$ — sublimation: 1 mol solid → gas directly
All three are positive, because energy must be supplied to separate the particles against the attractions holding them together.
Reverse any of them — freezing, condensing, depositing — and the sign becomes negative.

Sublimation is the sum of the other two

Enthalpy changes depend only on the start and end points, not the route taken. So going solid → gas directly must cost the same as going solid → liquid → gas.

$\Delta_{sub}H^\circ = \Delta_{fus}H^\circ + \Delta_{vap}H^\circ$

This is Hess's law applied to a change of state, and it is a common short exam question.

Why vaporisation costs far more than fusion

Melting only lets the particles move past one another. They stay close together, so most of the attractions between them remain.
Boiling completely separates the particles, so all the attractions between them must be overcome.
For water, $\Delta_{fus}H^\circ = 6.01$ kJ mol⁻¹ but $\Delta_{vap}H^\circ = 40.7$ kJ mol⁻¹ — nearly seven times as much.

The heating curve

Sloping sections — the temperature is rising, the substance is in one state, and the energy goes into making the particles move faster. Use $q = mc\Delta T$ .
Flat sections — the temperature does not change even though energy is still being supplied. Use the relevant $\Delta H$ of state change.
On a flat section the supplied energy is separating the particles, not speeding them up. The average kinetic energy is unchanged, so the thermometer does not move.
The flat section at the boiling point is much longer than the one at the melting point, because $\Delta_{vap}H$ is so much larger.

Multi-stage calculations

When a question spans a phase change, break it into separate stages and add them.
For each stage decide which relationship applies:
- temperature changing → $q = mc\Delta T$
- state changing → $n \times \Delta H$ for that change
Watch the units. $q = mc\Delta T$ works in grams and joules; the $\Delta H$ values are in kJ per mole. Convert everything to kJ before adding.

Tips and Tricks

Say why the temperature stays constant during a phase change, in the right words: the energy supplied is used to overcome the attractions between the particles, not to increase their kinetic energy. Writing only "the energy is used to change state" rarely gets the mark.

Do not write that the covalent bonds break when a molecular substance boils. Only the attractions between molecules are overcome.

In a multi-stage question, label your stages and add them at the end. Mixing joules and kilojoules mid-calculation is the usual failure — decide on kJ early and stick to it.

$q = mc\Delta T$ cannot be used across a phase change, because $\Delta T$ is zero there and it would give an answer of zero energy. Each flat section needs its own $n \times \Delta H$ term.

If you are asked for $\Delta_{sub}H$ and given the other two, just add them — that is the whole question.

Worked ExampleHeating and boiling water

Calculate the total energy needed to convert 50.0 g of liquid water at 18.0 °C into steam at 100 °C.

$c(\text{water}) = 4.18$ J g⁻¹ °C⁻¹, $\Delta_{vap}H^\circ(\text{H}_2\text{O}) = 40.7$ kJ mol⁻¹, $M(\text{H}_2\text{O}) = 18.0$ g mol⁻¹

Step 1 — Identify the stages

There are two distinct stages, and they need different relationships:

Stage 1: heating the liquid water from 18.0 °C up to 100 °C. The temperature changes, so use $q = mc\Delta T$ .
Stage 2: boiling all of it at 100 °C. The temperature stays constant, so use $n \times \Delta_{vap}H$ .

Step 2 — Stage 1, heating the liquid

$\Delta T = 100 - 18.0 = 82.0 \text{ °C}$

$q_1 = mc\Delta T = 50.0 \times 4.18 \times 82.0 = 17138 \text{ J}$

Convert to kilojoules straight away, so both stages are in the same unit:

$q_1 = 17.138 \text{ kJ}$

Step 3 — Stage 2, boiling the liquid

$\Delta_{vap}H$ is quoted per mole, so first find the amount of water.

$n = \frac{m}{M} = \frac{50.0}{18.0} = 2.7778 \text{ mol}$

$q_2 = n \times \Delta_{vap}H = 2.7778 \times 40.7 = 113.06 \text{ kJ}$

Step 4 — Add the stages

$q_{total} = q_1 + q_2 = 17.138 + 113.06 = 130.20 \text{ kJ}$

Phase changes

Covers AS91390

The three enthalpy changes of state

$\Delta_{fus}H^\circ$ — fusion: 1 mol solid → liquid (melting)
$\Delta_{vap}H^\circ$ — vaporisation: 1 mol liquid → gas (boiling)
$\Delta_{sub}H^\circ$ — sublimation: 1 mol solid → gas directly
All three are positive, because energy must be supplied to separate the particles against the attractions holding them together.
Reverse any of them — freezing, condensing, depositing — and the sign becomes negative.

Sublimation is the sum of the other two

Enthalpy changes depend only on the start and end points, not the route taken. So going solid → gas directly must cost the same as going solid → liquid → gas.

$\Delta_{sub}H^\circ = \Delta_{fus}H^\circ + \Delta_{vap}H^\circ$

This is Hess's law applied to a change of state, and it is a common short exam question.

Why vaporisation costs far more than fusion

Melting only lets the particles move past one another. They stay close together, so most of the attractions between them remain.
Boiling completely separates the particles, so all the attractions between them must be overcome.
For water, $\Delta_{fus}H^\circ = 6.01$ kJ mol⁻¹ but $\Delta_{vap}H^\circ = 40.7$ kJ mol⁻¹ — nearly seven times as much.

The heating curve

Sloping sections — the temperature is rising, the substance is in one state, and the energy goes into making the particles move faster. Use $q = mc\Delta T$ .
Flat sections — the temperature does not change even though energy is still being supplied. Use the relevant $\Delta H$ of state change.
On a flat section the supplied energy is separating the particles, not speeding them up. The average kinetic energy is unchanged, so the thermometer does not move.
The flat section at the boiling point is much longer than the one at the melting point, because $\Delta_{vap}H$ is so much larger.

Multi-stage calculations

When a question spans a phase change, break it into separate stages and add them.
For each stage decide which relationship applies:
- temperature changing → $q = mc\Delta T$
- state changing → $n \times \Delta H$ for that change
Watch the units. $q = mc\Delta T$ works in grams and joules; the $\Delta H$ values are in kJ per mole. Convert everything to kJ before adding.

Tips and Tricks

Do not write that the covalent bonds break when a molecular substance boils. Only the attractions between molecules are overcome.

In a multi-stage question, label your stages and add them at the end. Mixing joules and kilojoules mid-calculation is the usual failure — decide on kJ early and stick to it.

$q = mc\Delta T$ cannot be used across a phase change, because $\Delta T$ is zero there and it would give an answer of zero energy. Each flat section needs its own $n \times \Delta H$ term.

If you are asked for $\Delta_{sub}H$ and given the other two, just add them — that is the whole question.

Worked ExampleHeating and boiling water

Calculate the total energy needed to convert 50.0 g of liquid water at 18.0 °C into steam at 100 °C.

$c(\text{water}) = 4.18$ J g⁻¹ °C⁻¹, $\Delta_{vap}H^\circ(\text{H}_2\text{O}) = 40.7$ kJ mol⁻¹, $M(\text{H}_2\text{O}) = 18.0$ g mol⁻¹

Step 1 — Identify the stages

There are two distinct stages, and they need different relationships:

Stage 1: heating the liquid water from 18.0 °C up to 100 °C. The temperature changes, so use $q = mc\Delta T$ .
Stage 2: boiling all of it at 100 °C. The temperature stays constant, so use $n \times \Delta_{vap}H$ .

Step 2 — Stage 1, heating the liquid

$\Delta T = 100 - 18.0 = 82.0 \text{ °C}$

$q_1 = mc\Delta T = 50.0 \times 4.18 \times 82.0 = 17138 \text{ J}$

Convert to kilojoules straight away, so both stages are in the same unit:

$q_1 = 17.138 \text{ kJ}$

Step 3 — Stage 2, boiling the liquid

$\Delta_{vap}H$ is quoted per mole, so first find the amount of water.

$n = \frac{m}{M} = \frac{50.0}{18.0} = 2.7778 \text{ mol}$

$q_2 = n \times \Delta_{vap}H = 2.7778 \times 40.7 = 113.06 \text{ kJ}$

Step 4 — Add the stages

$q_{total} = q_1 + q_2 = 17.138 + 113.06 = 130.20 \text{ kJ}$