Chemical Thermodynamics · Lesson 4 of 10
Reaction Enthalpy, Standard States, and Phase Changes
“Interpret reaction heat using specified states, then follow energy through melting, evaporation, cooling, and freezing.”
• Interpret reaction enthalpy using a balanced reaction and specified physical states. • Explain the meaning of standard pressure and specified temperature. • Distinguish fusion, vaporisation, sublimation, and their reverse changes. • Calculate heat for phase changes and multi-stage cooling or heating. • Connect phase-change enthalpy with intermolecular interactions.
A reaction needs a clearly defined energy change
When reactants become products, their total enthalpy changes. The difference can be measured as heat under suitable constant-pressure conditions. Its value is useful for deciding how much heating or cooling an industrial reaction needs. The balanced equation and physical states must be known because they specify the initial and final systems.
For methane combustion to liquid water, the product sum contains one molar enthalpy of carbon dioxide and two of liquid water. The reactant sum contains one of methane and two of oxygen. Using water vapour instead defines a different endpoint and therefore a different reaction enthalpy. Absolute individual molar enthalpies are not needed when measurable differences or consistent reference data are used.
The reaction enthalpy when participating substances are in their specified standard states at the stated temperature, written . For the pure substances considered here, standard pressure is 1 bar. A temperature, often 298 K, must still be specified.
The standard-state symbol does not automatically mean 298 K, and it does not mean 0 degrees Celsius. Pure liquid ethanol at 298 K and 1 bar is an example. The temperature is part of the description, because enthalpies change with temperature. The next lesson explains how elemental reference states provide a common zero for formation data.
Changing phase without raising temperature
Heat supplied to ice at its melting point can melt the ice while the temperature stays constant. The added energy changes molecular arrangements and interactions instead of simply increasing temperature. This explains why cannot calculate the entire heat of a process that crosses a phase change.
| Forward phase change | Molar enthalpy | Reverse process |
|---|---|---|
| Solid to liquid: fusion | Freezing has the opposite sign | |
| Liquid to gas: vaporisation | Condensation has the opposite sign | |
| Solid directly to gas: sublimation | Deposition has the opposite sign |
For example, melting one mole of ice near 273 K requires about 6.00 kJ using the chapter’s rounded data. Freezing the same amount releases about 6.00 kJ under the same conditions. Vaporising one mole of water at its boiling point requires about 40.79 kJ, while its vaporisation enthalpy at 298 K is about 44.01 kJ. Those values are not interchangeable because the temperatures differ.
Problem
Find the enthalpy change when 2.0 mol of water freezes at 273 K, using .
- 1.The supplied value describes melting, .
- 2.Freezing reverses that change, so its molar enthalpy is .
- 3.Multiply by the amount: .
- 4.At constant pressure the water releases 12.0 kJ to its surroundings.
What controls the energy needed?
A liquid’s molecules attract one another. To create a gas, they must become widely separated, so vaporisation needs energy. Strong hydrogen bonding helps explain why water needs more energy per mole to vaporise than a liquid such as acetone, whose intermolecular attractions differ. The comparison concerns interactions between molecules, not breaking water into hydrogen and oxygen atoms.
The following selected phase data preserve the range of substances in the chapter’s comparison: molecular substances, a hydrogen-bonded liquid, and an ionic solid. Read the fusion value at the listed melting temperature and the vaporisation value at the listed boiling temperature. Small differences among rounded datasets are normal; do not combine values from different temperatures without accounting for the change.
| Substance | Melting T / K | Fusion / kJ mol−1 | Boiling T / K | Vaporisation / kJ mol−1 |
|---|---|---|---|---|
| Nitrogen | 63.15 | 0.72 | 77.35 | 5.59 |
| Ammonia | 195.40 | 5.65 | 239.73 | 23.35 |
| Hydrogen chloride | 159.0 | 1.992 | 188.0 | 16.15 |
| Acetone | 177.8 | 5.72 | 329.4 | 29.1 |
| Carbon tetrachloride | 250.16 | 2.5 | 349.69 | 30.0 |
| Water | 273.15 | 6.01 | 373.15 | 40.79 |
| Sodium chloride | 1081.0 | 28.8 | 1665.0 | 170.0 |
| Benzene | 278.65 | 9.83 | 353.25 | 30.8 |
Sublimation bypasses the liquid phase. Dry ice is solid carbon dioxide and sublimes near 195 K at atmospheric pressure; the chapter uses about for this change. Naphthalene can sublime slowly and is assigned about in the source discussion. The rate of visible change and the molar energy required are different properties.
Problem
A film of water on a swimmer has mass 18.0 g. Find the heat needed to evaporate it at 298 K and the corresponding internal-energy change. Use .
- 1.With molar mass , the amount is .
- 2.The constant-pressure heat required is .
- 3.Treat the vapour as ideal and neglect the liquid pV contribution. Then .
- 4.At 298 K, , giving .
- 5.Evaporation can draw the required heat from the swimmer and surroundings, producing a cooling effect.
Following a process in stages
When a process changes both temperature and phase, choose intermediate states that separate these effects. A cooling step uses heat capacity; a freezing step uses fusion enthalpy with a reversed sign. Enthalpy is a state function, so adding the stage changes gives the total between the same endpoints.
Problem
One mole of liquid water at 100 degrees Celsius is cooled to liquid at 0 degrees Celsius and then frozen there. Use and .
- 1.This starting state is liquid water, not steam. For 18 g, the cooling change is .
- 2.Freezing one mole at 0 degrees Celsius gives .
- 3.Add the stages: .
- 4.Only condensed phases are involved, so the small pV correction is neglected and .
- 5.If the initial sample had been steam instead, condensation would be an additional step; leaving it out would change the problem.
A flat temperature during melting does not mean zero heat transfer. Also, heating liquid from one temperature to another and vaporising it are separate contributions. Specify the starting phase before choosing a calculation.
Quiz
What does the standard-state symbol specify for the pure substances used here?
What is the sign of the enthalpy change for freezing under the reverse conditions of melting?
Why can ice absorb heat without its temperature rising while it melts?
How much heat is needed to vaporise 0.50 mol of water at a temperature where its molar vaporisation enthalpy is 40.8 kJ/mol?
Which change is sublimation?
A process cools liquid water and then freezes it. How is its total enthalpy change found?
Practice Problems
- Write the products-minus-reactants expression for the enthalpy change of methane combustion to carbon dioxide gas and liquid water.
- Explain why standard enthalpy data require both a pressure convention and a stated temperature.
- Find the heat needed to melt 9.0 g of ice at 0 degrees Celsius, using 6.00 kJ/mol for fusion.
- For one mole of water cooled from 10 degrees Celsius to ice at −10 degrees Celsius, use liquid molar heat capacity 75.3 J mol−1 K−1, solid molar heat capacity 36.8 J mol−1 K−1, and fusion enthalpy 6.03 kJ/mol. Show all three stages.
- Explain the cooling felt when water evaporates from skin without confusing evaporation with bond dissociation.
- Compare the physical meanings of fusion, vaporisation, and sublimation enthalpies and explain why reversing a phase change reverses the sign.
Key Takeaways
A reaction enthalpy belongs to a balanced process with specified amounts and physical states. Standard pressure is 1 bar; the temperature of a standard-state dataset must still be stated. Phase changes need enthalpy even when temperature stays constant. Separate warming or cooling from phase changes, then add their signed enthalpy contributions.