A dissolution has H = -10\ kJ mol^-1 and S = +50\ J mol^-1K^-1. State the sign of G at any temperature. [1 point]

College Board AP 2021 (style)-style practice: Section I (multiple choice). A salt dissolves with H > 0. The dissolution is thermodynamically favorable only if (A) S 0 and T S > H (C) the temperature is very low (D) K_sp is very small. Justify your choice.

A 1-point conceptual MCQ. The answer is (B). With H > 0, the process is favorable (G 0 and T S exceeds H. The trap is (C): low temperature makes the entropy term smaller, so it would make dissolution less favorable, not more.

United StatesChemistrySyllabus dot point

How does the free energy of dissolution relate to whether and how much a salt dissolves?

Topic 7.14 Free Energy of Dissolution: relate the thermodynamic favourability of dissolving a salt to the enthalpy and entropy of dissolution and to the sign of the free energy change.

A focused answer to AP Chemistry Topic 7.14, covering the enthalpy and entropy of dissolution, how their balance sets the free energy of dissolution, and how the sign of the free energy change relates to solubility, with full worked examples.

Generated by Claude Opus 4.810 min answerUpdated 2026-06-04

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

Whether a salt dissolves is governed by the free energy of dissolution, $\Delta G = \Delta H - T\Delta S$ . The enthalpy of dissolution can be positive (endothermic, the solution cools) or negative (exothermic, the solution warms), depending on whether breaking the lattice costs more or less energy than hydrating the ions releases. The entropy of dissolution is usually positive, because an ordered solid lattice becomes dispersed, freely moving hydrated ions, a more disordered state. A salt dissolves to a significant extent when the free energy change is negative; many salts dissolve even though the process is endothermic, because the favorable entropy term outweighs the unfavorable enthalpy. The sign of the overall free energy change therefore depends on the balance of enthalpy and entropy, and the temperature scales the entropy contribution.

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What this topic is asking
Enthalpy of dissolution
Entropy of dissolution
Free energy decides
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What this topic is asking

The College Board (Topic 7.14) wants you to relate the thermodynamic favourability of dissolving a salt to the enthalpy and entropy of dissolution and to the sign of the free energy change. This connects solubility (an equilibrium idea) to thermodynamics (Unit 9), previewing how $\Delta G$ governs whether a process happens.

Enthalpy of dissolution

So whether dissolving heats or cools the solution depends on the competition between lattice breaking and ion hydration. A cold pack (ammonium nitrate) is endothermic; a hot pack (some other salts) is exothermic. The enthalpy alone does not decide whether the salt dissolves.

Entropy of dissolution

Breaking up the regular lattice and spreading the ions through the solvent greatly increases the disorder of the system, so dissolution is typically entropy-favored. This is why many salts dissolve readily even when the process absorbs heat: the entropy gain drives them.

Free energy decides

The sign of the free energy of dissolution decides favourability:

\Delta G_\text{soln} = \Delta H_\text{soln} - T\Delta S_\text{soln}

A salt dissolves to a significant extent when $\Delta G_\text{soln} < 0$ . With a positive $\Delta S$ , the $-T\Delta S$ term is negative and favors dissolution, and it grows with temperature. So even an endothermic dissolution ( $\Delta H > 0$ ) is favorable if the favorable entropy term is large enough to outweigh it. This balance, and the sign of $\Delta G$ , links directly to the equilibrium constant ( $K_\text{sp}$ ) through the relationship developed in Unit 9.

Reasoning about an endothermic dissolution

A salt dissolves with $\Delta H_\text{soln} = +25\ \text{kJ mol}^{-1}$ and $\Delta S_\text{soln} = +120\ \text{J mol}^{-1}\text{K}^{-1}$ at $298$ K. Is dissolution favorable?

step 1 Identify the signs

$\Delta H > 0$ (unfavorable), $\Delta S > 0$ (favorable).

step 2 Compute the entropy term

$T\Delta S = (298)(120) = 35760\ \text{J mol}^{-1} = 35.8\ \text{kJ mol}^{-1}$ .

step 3 Compute the free energy

$\Delta G = \Delta H - T\Delta S = 25 - 35.8 = -10.8\ \text{kJ mol}^{-1}$ .

step 4 Conclude

$\Delta G < 0$ , so the dissolution is thermodynamically favorable; the entropy gain outweighs the endothermic enthalpy, and the salt dissolves despite cooling the solution.

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Q1. State the usual sign of the entropy of dissolution of a salt and explain. [2 points]

Cue. Positive, because the ordered solid lattice becomes dispersed, disordered hydrated ions in solution.

Q2. A dissolution has $\Delta H = -10\ \text{kJ mol}^{-1}$ and $\Delta S = +50\ \text{J mol}^{-1}\text{K}^{-1}$ . State the sign of $\Delta G$ at any temperature. [1 point]

Cue. Negative at all temperatures (favorable $\Delta H$ and favorable $\Delta S$ both make $\Delta G < 0$ ).

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

AP 2023 (style)4 marksSection II (long FRQ, part). When

\text{NH}_4\text{NO}_3

dissolves in water, the solution gets colder, yet the salt dissolves readily. (a) State the sign of

\Delta H

for the dissolution and justify from the temperature change. (b) Explain how the salt can dissolve despite an unfavorable enthalpy. (c) State the sign of

\Delta S

for the dissolution and justify. (d) Relate the sign of

\Delta G

to the fact that the salt dissolves.

Show worked answer →

A 4-point conceptual FRQ on the free energy of dissolution.

(a) Enthalpy sign (1 point): the solution cools, so the process absorbs heat and is endothermic; $\Delta H > 0$ .
(b) Despite unfavorable enthalpy (1 point): the dissolution is driven by entropy; the ordered solid breaks up into freely moving hydrated ions, increasing disorder, and the positive $\Delta S$ term ( $-T\Delta S$ is negative) outweighs the positive $\Delta H$ .
(c) Entropy sign (1 point): $\Delta S > 0$ , because the solid lattice becomes dispersed ions in solution, a more disordered arrangement.
(d) Free energy (1 point): the salt dissolves, so the process is thermodynamically favorable and $\Delta G < 0$ ; here $\Delta G = \Delta H - T\Delta S$ is negative because the favorable entropy term dominates the unfavorable enthalpy.

Markers reward the positive $\Delta H$ from cooling, the entropy-driven reasoning, the positive $\Delta S$ , and the negative $\Delta G$ for a salt that dissolves.

AP 2021 (style)1 marksSection I (multiple choice). A salt dissolves with

\Delta H > 0

. The dissolution is thermodynamically favorable only if (A)

\Delta S < 0

(B)

\Delta S > 0

and

T\Delta S > \Delta H

K_\text{sp}

is very small. Justify your choice.

Show worked answer →

A 1-point conceptual MCQ. The answer is (B).

With $\Delta H > 0$ , the process is favorable ( $\Delta G < 0$ ) only if the entropy term $-T\Delta S$ is sufficiently negative, that is, $\Delta S > 0$ and $T\Delta S$ exceeds $\Delta H$ . The trap is (C): low temperature makes the entropy term smaller, so it would make dissolution less favorable, not more.

Related dot points

Sources & how we know this

AP Chemistry Course and Exam Description — College Board (2020)