A dissolution has H = -5.0\ kJ mol^-1 and S = +40\ J mol^-1K^-1. State the sign of G at all temperatures. [1 point]

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How do enthalpy, entropy and free energy together explain the thermodynamics of dissolving a salt?

Topic 9.6 Free Energy of Dissolution: analyze the dissolution of a salt using delta G equals delta H minus T delta S, and relate the sign of delta G to whether and how much the salt dissolves.

A focused answer to AP Chemistry Topic 9.6, covering the thermodynamics of dissolution, how the enthalpy and entropy of solution combine into the free energy, and how the sign of delta G relates to solubility and Ksp, with full worked examples.

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

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Quick answer

Dissolving a salt is governed by the free energy of dissolution, delta G equals delta H minus T delta S. The enthalpy of dissolution is the net of breaking the ionic lattice (endothermic) and hydrating the ions (exothermic), and can be positive or negative. The entropy of dissolution is usually positive, because the ordered solid disperses into freely moving hydrated ions. The salt dissolves to a significant extent when delta G is negative. Many salts dissolve even though dissolving is endothermic, because the favorable entropy term outweighs the unfavorable enthalpy, and the entropy contribution grows with temperature, so some salts dissolve only when warm. Through the relationship delta G standard equals minus RT ln K, a negative free energy of dissolution corresponds to a larger solubility product, linking the thermodynamics directly to Ksp.

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What this topic is asking
The thermodynamics of dissolving
Endothermic salts that still dissolve
Linking to the solubility product
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What this topic is asking

The College Board (Topic 9.6) wants you to analyze the dissolution of a salt using $\Delta G = \Delta H - T\Delta S$ , and to relate the sign of $\Delta G$ to whether and how much the salt dissolves. This applies the Gibbs free energy framework of Topic 9.3 specifically to dissolving, deepening the solubility ideas of Unit 7.

The thermodynamics of dissolving

So dissolving is decided by the same balance of enthalpy and entropy as any other process. Because the entropy of dissolution is typically positive, the $-T\Delta S$ term favors dissolving, and this contribution grows with temperature. The enthalpy can help or hinder, depending on whether hydration releases more energy than the lattice costs.

Endothermic salts that still dissolve

A salt can dissolve readily even when dissolving absorbs heat ( $\Delta H_\text{soln} > 0$ ), provided the favorable entropy term is large enough to make $\Delta G < 0$ . This is why ammonium nitrate dissolves and cools the solution: the process is endothermic, but the large entropy gain from dispersing the ions drives it. The temperature dependence means some sparingly soluble salts dissolve more as the solution is warmed, because the $-T\Delta S$ term becomes more negative.

Linking to the solubility product

Through the relationship $\Delta G^\circ = -RT\ln K_\text{sp}$ (Topic 9.5), the sign of the free energy of dissolution determines the size of the solubility product. A negative $\Delta G^\circ$ corresponds to a larger $K_\text{sp}$ (more soluble); a positive $\Delta G^\circ$ corresponds to a small $K_\text{sp}$ (sparingly soluble). So the thermodynamic analysis of dissolution and the equilibrium description by $K_\text{sp}$ are two views of the same thing, connected by the free-energy-equilibrium equation.

Free energy of an endothermic dissolution

A salt has $\Delta H_\text{soln} = +12.0\ \text{kJ mol}^{-1}$ and $\Delta S_\text{soln} = +60.0\ \text{J mol}^{-1}\text{K}^{-1}$ . Find $\Delta G$ at $298$ K and the crossover temperature.

step 1 Convert units and apply the equation

$\Delta S_\text{soln} = 0.0600\ \text{kJ mol}^{-1}\text{K}^{-1}$ ; $\Delta G = \Delta H - T\Delta S = 12.0 - (298)(0.0600)$ .

step 2 Evaluate at 298 K

$\Delta G = 12.0 - 17.9 = -5.9\ \text{kJ mol}^{-1}$ , so the salt dissolves at $298$ K.

step 3 Find the crossover temperature

$\Delta G = 0$ when $T = \dfrac{\Delta H}{\Delta S} = \dfrac{12.0}{0.0600} = 200$ K.

step 4 Interpret

Above $200$ K the dissolution is favorable; at room temperature ( $298$ K) it dissolves despite being endothermic, driven by the positive entropy change.

Try this

Q1. A dissolution has $\Delta H = -5.0\ \text{kJ mol}^{-1}$ and $\Delta S = +40\ \text{J mol}^{-1}\text{K}^{-1}$ . State the sign of $\Delta G$ at all temperatures. [1 point]

Cue. Negative at all temperatures (favorable enthalpy and favorable entropy), so the salt dissolves readily.

Q2. Explain why warming a solution can increase the solubility of an endothermic salt. [2 points]

Cue. The dissolution has a positive entropy change, so raising the temperature makes $-T\Delta S$ more negative, lowering $\Delta G$ and favoring dissolving.

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). A salt has dissolution values

\Delta H = +18.0\ \text{kJ mol}^{-1}

and

\Delta S = +75.0\ \text{J mol}^{-1}\text{K}^{-1}

. (a) Calculate

\Delta G

for dissolution at

298

K. (b) State whether the salt dissolves to a significant extent at

298

K, and justify. (c) Determine the temperature above which dissolution becomes favorable. (d) Relate the sign of

\Delta G

to whether

K_\text{sp}

is greater or less than 1.

Show worked answer →

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

(a) $\Delta G$ (1 point): $\Delta S = 0.0750\ \text{kJ mol}^{-1}\text{K}^{-1}$ ; $\Delta G = 18.0 - (298)(0.0750) = 18.0 - 22.4 = -4.4\ \text{kJ mol}^{-1}$ .
(b) Dissolves (1 point): $\Delta G < 0$ at $298$ K, so the dissolution is thermodynamically favorable and the salt dissolves to a significant extent.
(c) Crossover temperature (1 point): favorable when $\Delta G < 0$ , i.e. $\Delta H < T\Delta S$ ; $T > \dfrac{18.0}{0.0750} = 240$ K, so it is favorable above $240$ K (and at $298$ K).
(d) Ksp relation (1 point): $\Delta G < 0$ corresponds (via $\Delta G^\circ = -RT\ln K$ ) to $K_\text{sp} > 1$ for this idealized case, meaning a relatively soluble salt; a positive $\Delta G$ would give $K_\text{sp} < 1$ .

Markers reward the $\Delta G$ value, the favorable conclusion, the crossover temperature, and the link to the size of $K_\text{sp}$ .

AP 2021 (style)1 marksSection I (multiple choice). A salt dissolves endothermically yet readily. This is best explained by (A) a negative entropy change (B) a positive entropy change that makes

\Delta G

negative (C) a very small

K_\text{sp}

(D) a catalyst. Justify your choice.

Show worked answer →

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

An endothermic dissolution ( $\Delta H > 0$ ) is favorable only if the entropy term overcomes it; a positive $\Delta S$ makes $-T\Delta S$ negative enough that $\Delta G < 0$ , so the salt dissolves. The trap is (A): a negative entropy change would make dissolution less favorable, not more.

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Sources & how we know this

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