AP® Chemistry review sheet from Aim for Five (aimforfive.com/chem/units/9/9-6)
Unit 9 · Topic 9.6
9.6 Free Energy of Dissolution
Dissolving a solid involves three steps: breaking apart the solid, making room in the solvent, and forming new attractions between the dissolved particles and the solvent. Each step has its own enthalpy and entropy change, and because they partly cancel, predicting whether ΔG° for dissolving is negative can be tricky.
Key terms
- dissolution
- ion-dipole attraction
- hydration
- enthalpy change
- entropy change
Three steps of dissolving
Think of dissolving an ionic solid in water as three steps, even though they really happen together.
| Step | What happens | Enthalpy | Entropy |
|---|---|---|---|
| 1. Break up the solid | Ions (or molecules) are separated from the crystal | Positive (absorbs energy) | Positive (particles spread out) |
| 2. Make room in the solvent | Some attractions between water molecules are disrupted | Positive (absorbs energy) | Usually positive |
| 3. Solvate the particles | Water molecules cluster around ions (ion–dipole attractions) | Negative (releases energy) | Negative (water becomes more ordered around ions) |
Why the result is hard to predict
The overall ΔH of dissolving is the sum of a large positive term (steps 1 and 2) and a large negative term (step 3). The difference between two large numbers can be small and of either sign, so dissolving can be exothermic or endothermic (topic 6.1).
The overall ΔS has the same problem. Breaking up the solid increases entropy, but organizing water molecules around the ions decreases it. Small or highly charged ions, such as Mg²⁺ or Ca²⁺, hold water molecules tightly and can make ΔS of dissolving negative.
ΔG° = ΔH° − TΔS° then combines two quantities that are each the result of a partial cancellation. You can usually say whether each step raises or lowers H and S, and roughly by how much. Adding them all up into a reliable ΔG° is much harder, because the big terms nearly cancel.
Real examples
- NH₄NO₃: dissolving is endothermic (about +26 kJ/mol) but has a large entropy increase (about +110 J/(mol·K)), so ΔG° is negative (about −7 kJ/mol at 25 °C). It dissolves readily, and the solution gets cold. That's how instant cold packs work.
- NaCl: ΔH of dissolving is small and positive (about +4 kJ/mol). A modest entropy increase is enough to make ΔG° negative, so NaCl is quite soluble.
- CaCl₂ and NaOH: dissolving is strongly exothermic. That's why CaCl₂ is used in some hot packs and why dissolving NaOH pellets warms the water a lot.
Linking ΔG° to solubility
For a salt dissolving, the equilibrium constant is Ksp. By ΔG° = −RT ln K (topic 9.5), a salt with a negative ΔG° of dissolving has Ksp > 1, which is what the solubility rules call soluble (topic 7.11). A salt with a large positive ΔG° has a tiny Ksp.
Putting it together
When you analyze any dissolving process, check the signs step by step: separating particles absorbs energy and spreads matter out, while solvation releases energy but organizes the solvent. The observed temperature change gives you the sign of the overall ΔH, and whether the solid dissolves tells you about ΔG.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Explaining an endothermic dissolving
When NH₄NO₃ dissolves in water, the temperature drops, yet the salt is very soluble. Explain using enthalpy and entropy, and identify the signs of ΔH°, ΔS° and ΔG° for dissolving.
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- Step 1: The temperature drops, so dissolving absorbs energy from the water: ΔH° > 0. The energy needed to separate the ions and disrupt water–water attractions is greater than the energy released when water molecules surround the NH₄⁺ and NO₃⁻ ions.
- Step 2: The salt dissolves readily, so dissolving must be thermodynamically favored: ΔG° < 0.
- Step 3: Since ΔG° = ΔH° − TΔS° and ΔH° is positive, ΔG° can be negative only if TΔS° is larger than ΔH°. So ΔS° must be positive and large enough.
- Step 4: The positive ΔS° comes mainly from the ions leaving the ordered crystal and dispersing through the solution.
Answer: ΔH° > 0, ΔS° > 0 (large), ΔG° < 0: the entropy increase drives the dissolving even though it absorbs energy.
- Example 2
Trap: one factor isn't the whole story
A student predicts that a salt will dissolve because the ion–dipole attractions it forms with water release a lot of energy. Explain why that alone isn't enough to make the prediction.
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- Step 1: Strong ion–dipole attractions make step 3 very exothermic, but steps 1 and 2 absorb energy. If the ions are held very strongly in the crystal (for example, small ions with large charges), step 1 can absorb even more.
- Step 2: Strong ion–dipole attractions also order the surrounding water molecules, which lowers entropy and can make the overall ΔS negative.
- Step 3: So a large energy release in one step doesn't guarantee a negative overall ΔH, and even a negative ΔH doesn't guarantee a negative ΔG if ΔS is negative.
- Step 4: You need to weigh all three steps for both enthalpy and entropy. That's why predicting solubility from structure alone is difficult.
Answer: One step can't decide it: the energy to break up the solid and the entropy cost of ordering water can cancel or outweigh the energy released by ion–dipole attractions.
Common mistakes
- Assuming every endothermic dissolving process is unfavorable.
- Assuming ΔS of dissolving is always positive. Ordering of water around small, highly charged ions can make it negative.
- Considering only one step (usually the ion–dipole attractions) when predicting whether a salt dissolves.
- Getting the sign of each step backward: separating particles absorbs energy; forming attractions releases it.
On the exam
- Expect questions that give an observation (temperature change and whether a salt dissolves) and ask you to deduce the signs of ΔH°, ΔS° and ΔG°.
- When explaining, name the interactions broken (ion–ion, water–water) and formed (ion–dipole), and connect them to the signs.
Connected topics
Videos
Check yourself
4 questions on 9.6 Free Energy of Dissolution. Pick an answer to see if you got it, and why.
NH₄NO₃(s) dissolves readily in water at 25 °C, and the solution gets colder. Which of the following best explains why the dissolving is thermodynamically favored?
Dissolving an ionic solid in water can be thought of as three steps. Which step releases energy?
For NaCl(s) → Na⁺(aq) + Cl⁻(aq), ΔH° = +3.9 kJ/mol and ΔS° = +43 J/(mol·K). What is ΔG° at 298 K, and what drives the dissolving?
When an ionic solid dissolves in water, which of the following steps tends to decrease the entropy of the system?
0 of 4 answered