Free Energy of Dissolution: Study Guide (Chemistry 9.6)

Dissolving as a thermodynamic process

In Topic 3.10 you explained solubility with intermolecular forces and “like dissolves like”. In Topic 7.11 you treated dissolving a sparingly soluble salt as an equilibrium with constant K_sp. Now you can join these with free energy.

For a salt MX dissolving in water:

MX(s) ⇌ M⁺(aq) + X⁻(aq) K = K_sp

The free energy change for this process, ΔG°(dissolution), works like any other ΔG°:

• ΔG° = ΔH° − TΔS° (Topic 9.3), and
• ΔG° = −RT ln K_sp (Topic 9.5).

So a salt with a negative ΔG° of dissolution has K_sp > 1 and is very soluble. A salt with a large positive ΔG° has a tiny K_sp and is only sparingly soluble. The question for this topic is: what decides ΔH° and ΔS° for dissolving, and why is the result so hard to predict?

Three factors at the particle level

The CED describes dissolving in terms of three factors. Each one has an enthalpy part and an entropy part.

Factor	What happens to the particles	ΔH contribution	ΔS contribution
1. Breaking up the solid	Ions are pulled apart, overcoming the ion–ion attractions that hold the lattice together	Positive and large: attractions are overcome	Positive: ions leave a fixed, ordered lattice and can spread through the whole solution
2. Reorganising the solvent	Some water–water hydrogen bonds are broken to make room, and water molecules turn to face the ions in shells	Usually positive: some solvent–solvent attractions are lost	Usually negative: water molecules near the ions are held in more ordered arrangements
3. Ion–solvent attractions	Each ion is surrounded by water molecules attracted to it (ion–dipole)	Negative and large: new attractions form	Closely tied to factor 2: the same attractions hold water molecules in place

Two patterns matter most:

• Enthalpy. Factor 1 costs a lot of energy; factor 3 gives back a lot. ΔH° of dissolving is the small difference between two large numbers. It can be positive (endothermic dissolving) or negative (exothermic dissolving).
• Entropy. Factor 1 increases entropy; factors 2 and 3 decrease it. Again the total can go either way.

How the size and charge of the ions matter

• Small or highly charged ions (for example 2+, 3+ or 2− ions) attract each other strongly in the solid, so factor 1 needs more energy. They also attract water strongly, so factor 3 releases more energy. In addition, they hold more water molecules more tightly, so the ordering in factor 2 is larger and the entropy change of dissolving is less positive, or even negative.
• Large ions with a single charge attract water less strongly. Fewer water molecules are ordered, so the entropy gain from freeing the ions usually wins: ΔS° of dissolving is positive.

This is one reason why many salts made of two highly charged ions, such as many carbonates and phosphates with 2+ or 3+ metal ions, are only sparingly soluble.

Why predictions are hard: cancellation

The CED stresses one idea: you can estimate the sign and relative size of each contribution, but predicting the total ΔG° of dissolution is difficult. The reason is cancellation.

• In enthalpy, a large positive term (factor 1) is nearly cancelled by a large negative term (factor 3). A small percentage error in either can flip the sign of ΔH°.
• In entropy, the positive term (freeing the ions) is partly cancelled by the negative term (ordering the solvent).
• Then ΔH° and −TΔS° can themselves have opposite signs and partly cancel again.

So the particle picture explains why a salt behaves as it does once you know the data. It is much weaker at predicting the result in advance. This is the “degree to which a model describes” a system that the course asks you to judge.

Temperature and solubility

Because ΔG° = ΔH° − TΔS°, temperature changes ΔG° of dissolution and therefore K_sp.

• If dissolving is endothermic (ΔH° > 0), K_sp increases as the temperature rises: the salt becomes more soluble.
• If dissolving is exothermic (ΔH° < 0), K_sp decreases as the temperature rises.

This follows from K = e^(−ΔH°/RT + ΔS°/R), assuming ΔH° and ΔS° are roughly constant: only the ΔH° term depends on T. It matches Le Chatelier’s principle, with heat treated as a reactant for an endothermic process.

Worked example 1: an endothermic salt that still dissolves

Question. For the invented salt MX, ΔH°(dissolution) = +18.0 kJ mol⁻¹ and ΔS°(dissolution) = +70.0 J mol⁻¹ K⁻¹. (a) Calculate ΔG° and K_sp at 298 K. (b) Explain the signs of ΔH° and ΔS° at the particle level. (c) Predict, with a calculation, how K_sp changes at 323 K.

(a)

1. ΔG° = ΔH° − TΔS° = 18.0 − (298)(0.0700) = 18.0 − 20.86 = −2.86 kJ mol⁻¹.
2. K_sp = e^(−ΔG°/RT) = e^(2860 / 2477.6) = e^1.154 = 3.17.

ΔG° is negative and K_sp > 1, so dissolving is thermodynamically favoured: MX is very soluble, even though dissolving it absorbs heat.

(b) ΔH° is positive, so the energy needed to separate the ions (factor 1) and to break some water–water attractions (factor 2) is slightly more than the energy released by the new ion–dipole attractions (factor 3). ΔS° is positive, so the entropy gained by freeing the ions from the lattice is larger than the entropy lost as water molecules are ordered around them. This suggests the ions are not small or highly charged.

(c) At 323 K: ΔG° = 18.0 − (323)(0.0700) = 18.0 − 22.61 = −4.61 kJ mol⁻¹. RT = 8.314 × 323 = 2685.4 J mol⁻¹, so K_sp = e^(4610 / 2685.4) = e^1.717 = 5.57. K_sp increases, so MX is more soluble in warm water, as expected for endothermic dissolving.

Worked example 2: exothermic, yet barely soluble

Question. Two invented 1:1 salts are compared at 298 K.

Salt	Ions	ΔH°(dissolution)	ΔS°(dissolution)
A	large, charges +1 and −1	+5.0 kJ mol⁻¹	+40 J mol⁻¹ K⁻¹
B	small, charges +2 and −2	−10.0 kJ mol⁻¹	−120 J mol⁻¹ K⁻¹

(a) Calculate ΔG° and K_sp for each salt. (b) Explain why B, which dissolves exothermically, is far less soluble than A.

(a)

• Salt A: ΔG° = 5.0 − (298)(0.040) = 5.0 − 11.92 = −6.92 kJ mol⁻¹; K_sp = e^(6920/2477.6) = e^2.793 = 16.3.
• Salt B: ΔG° = −10.0 − (298)(−0.120) = −10.0 + 35.76 = +25.8 kJ mol⁻¹; K_sp = e^(−25 760/2477.6) = e^(−10.40) = 3.05 × 10⁻⁵.

(b) Enthalpy alone would wrongly suggest that B is the more soluble salt. The difference is entropy. B’s small, doubly charged ions attract water molecules very strongly and hold many of them in ordered shells (factors 2 and 3). This ordering outweighs the entropy gained by freeing the ions, so ΔS° is strongly negative. At 298 K, −TΔS° = +35.8 kJ mol⁻¹, which more than cancels the favourable ΔH° of −10.0 kJ mol⁻¹. For A, the large singly charged ions order fewer water molecules, ΔS° is positive, and −TΔS° = −11.9 kJ mol⁻¹ outweighs the small positive ΔH°.

Check. The difference in ΔG° is 25.8 − (−6.9) = 32.7 kJ mol⁻¹. At 298 K each 5.7 kJ mol⁻¹ is a factor of 10, and 32.7 ÷ 5.7 ≈ 5.7, so the K_sp values should differ by about 10⁵·⁷ ≈ 5 × 10⁵. They do: 16.3 ÷ (3.05 × 10⁻⁵) ≈ 5 × 10⁵.

Worked example 3: how far does the enthalpy-only model go?

Question. Dissolving ammonium nitrate, NH₄NO₃, in water is strongly endothermic: the solution becomes noticeably colder, which is why it is used in some instant cold packs. A student says: “Since dissolving NH₄NO₃ absorbs heat, it cannot be thermodynamically favoured.” Explain the degree to which an enthalpy-only model describes this process.

Answer.

1. What the model gets right. The positive ΔH° correctly tells you that the attractions broken (ion–ion in the solid, some water–water hydrogen bonds) need more energy than the new ion–dipole attractions release. This explains why the solution cools.
2. Where it fails. NH₄NO₃ in fact dissolves readily in water. So ΔG° of dissolution must be negative. Since ΔH° > 0, the only way this can happen is a positive ΔS° large enough that TΔS° > ΔH°.
3. Particle-level reason. The large, singly charged ions NH₄⁺ and NO₃⁻ leave an ordered lattice and spread through the solution, a large entropy gain. They order relatively few water molecules, so the entropy loss from factor 2 is smaller.
4. Conclusion. An enthalpy-only model explains the temperature change but not the solubility. You need both ΔH° and ΔS° (through ΔG°) to explain whether a salt dissolves.

Common misconceptions

• “Endothermic dissolving means the salt is insoluble.” Not if ΔS° is positive and large enough. Cold-pack salts dissolve readily.
• “Exothermic dissolving means the salt is soluble.” Not if ΔS° is strongly negative, as for many salts of small, highly charged ions (Worked example 2).
• “Dissolving always increases entropy.” Freeing ions increases entropy, but ordering water around ions decreases it. The total can be negative.
• “Breaking the lattice releases energy.” Separating ions always needs energy. Energy is released when the ions are attracted to water molecules.
• “Dissolving breaks covalent bonds inside polyatomic ions.” NH₄⁺, NO₃⁻ and SO₄²⁻ stay intact; only the attractions between ions are overcome.
• “A positive ΔG° of dissolution means none of the salt dissolves.” It means K_sp < 1. A small amount always dissolves.
• “Hot water always dissolves more salt.” That is true only when dissolving is endothermic. When it is exothermic, K_sp falls as temperature rises.

Where this leads

This topic used free energy to explain a physical process. Next, Topic 9.7, Coupled Reactions, shows how a reaction with a positive ΔG° can be driven by joining it to one with a large negative ΔG°. Try the practice questions now, then use the revision notes and the checklist to consolidate.