Free Energy of Dissolution溶解的自由能

Big Idea · Scale, Proportion, and Quantity (SPQ)AP Science Practice 6 · ArgumentationAP link · Unit 7~40 min

Learning goals

  • Describe the enthalpy change of dissolving in three steps (separating solute particles, separating solvent molecules, and solute–solvent attraction), and describe the entropy change of dissolving
  • Use ΔG° = ΔH° − TΔS° to decide whether dissolving is thermodynamically favorable, and explain how endothermic dissolving can still be spontaneous
  • Use ΔG° = −RT ln Ksp to connect the free energy of dissolution with the solubility product

Before you start

  • The dissolving process and “like dissolves like” (2.9.1)
  • Endothermic and exothermic processes (U6.1)
  • Introduction to solubility equilibria (Ksp) (U7.11)

The enthalpy of dissolving in three steps

Dissolving an ionic compound can be pictured in three steps (U6.2, 2.9.1):

  1. separate the ions in the crystal, overcoming their attraction: endothermic;
  2. push water molecules apart to make room for the ions, overcoming hydrogen bonds: endothermic;
  3. let the ions attract water molecules (ion–dipole attraction, hydration): exothermic.

The sum of the three steps is ΔH° of dissolving. If step 3 releases more energy than the first two absorb, dissolving is exothermic (NaOH, for example); if less, it is endothermic (NH₄NO₃ or KCl, for example).

The entropy of dissolving

When ions spread out from an orderly crystal into solution, disorder increases, so entropy usually increases (ΔS° > 0). But water molecules attracted to each ion line up around it more orderly, which lowers the entropy. Small, highly charged ions are strongly hydrated, and this decrease can be large: in OpenStax Appendix G, S° for Mg²⁺(aq) is −138.1 J/(K·mol), a negative value.

Free energy decides

Whether a process is thermodynamically favorable at a given temperature depends on the Gibbs free energy change吉布斯自由能变, ΔG:

ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

Dissolving is favorable when ΔG° < 0 and unfavorable when ΔG° > 0 (Unit 9 covers this in detail). It is related to the equilibrium constant by ΔG° = −RT ln K, and for a slightly soluble salt K is Ksp: the more positive ΔG°, the smaller Ksp.

Endothermic dissolving can still be spontaneous, as long as the entropy increase is large enough to make −TΔS° more negative than ΔH° is positive. Choose a salt and drag the temperature to watch the three bars:

Try it: the free energy of dissolution

Example 1: Why NaCl is soluble

From OpenStax Appendix G (ΔfH° in kJ/mol, S° in J/(K·mol)): NaCl(s) −411.2, 72.1; Na⁺(aq) −240.1, 59; Cl⁻(aq) −167.2, 56.5. Find ΔH°, ΔS°, and ΔG° at 25 °C for NaCl(s) → Na⁺(aq) + Cl⁻(aq).

Step-by-step solution

  1. Formulas. Both are “products minus reactants,” each term times its coefficient (all 1 here): ΔH° = ΣΔfH°(products) − ΣΔfH°(reactants) and ΔS° = ΣS°(products) − ΣS°(reactants).

  2. Calculate ΔH° (in kJ/mol).

    ΔH∘=[(−240.1)+(−167.2)]−(−411.2)=−407.3+411.2=+3.9\begin{aligned} \Delta H^\circ &= [(-240.1) + (-167.2)] \\ &\quad - (-411.2) \\ &= -407.3 + 411.2 \\ &= +3.9 \end{aligned}

    Dissolving is slightly endothermic.

  3. Calculate ΔS° (in J/(K·mol)).

    ΔS∘=(59+56.5)−72.1=+43.4\begin{aligned} \Delta S^\circ &= (59 + 56.5) - 72.1 \\ &= +43.4 \end{aligned}

    The ions spread out into the water, so entropy increases.

  4. Make the units match. ΔH° is in kJ but ΔS° is in J, so convert ΔS° to kJ before multiplying:

    ΔS∘=+43.4 J/(K⋅mol)=+0.0434 kJ/(K⋅mol)\begin{aligned} \Delta S^\circ &= +43.4\ \text{J/(K·mol)} \\ &= +0.0434\ \text{kJ/(K·mol)} \end{aligned}

  5. Calculate TΔS° and ΔG° (in kJ/mol). T = 298.15 K:

    TΔS∘=298.15×0.0434=12.9ΔG∘=ΔH∘−TΔS∘=3.9−12.9=−9.0\begin{aligned} T\Delta S^\circ &= 298.15 \times 0.0434 = 12.9 \\ \Delta G^\circ &= \Delta H^\circ - T\Delta S^\circ \\ &= 3.9 - 12.9 = \boxed{-9.0} \end{aligned}

  6. Conclusion. ΔG° < 0, so dissolving is favorable under standard conditions. ΔH° is positive (unfavorable), but −TΔS° = −12.9 kJ/mol is more negative than ΔH° is positive, so dissolving NaCl is driven by entropy.

Example 2: Ksp from ΔG°

From the free energies of formation in OpenStax Appendix G, ΔG° at 25 °C for AgCl(s) → Ag⁺(aq) + Cl⁻(aq) is +55.7 kJ/mol. Find Ksp for AgCl.

Step-by-step solution

  1. Formula. For a solubility equilibrium, K is Ksp:

    ΔG∘=−RTln⁡Kspln⁡Ksp=−ΔG∘RT\begin{gathered} \Delta G^\circ = -RT \ln K_{sp} \\ \ln K_{sp} = -\frac{\Delta G^\circ}{RT} \end{gathered}

  2. Make the units match. R = 8.314 J/(K·mol) is in joules, so ΔG° must be in joules too:

    ΔG∘=+55.7 kJ/mol=+55 700 J/mol\begin{aligned} \Delta G^\circ &= +55.7\ \text{kJ/mol} \\ &= +55\,700\ \text{J/mol} \end{aligned}

  3. Calculate RT.

    RT=8.314×298.15=2479 J/mol\begin{aligned} RT &= 8.314 \times 298.15 \\ &= 2479\ \text{J/mol} \end{aligned}

  4. Calculate ln Ksp.

    ln⁡Ksp=−55 7002479=−22.47\ln K_{sp} = -\frac{55\,700}{2479} = -22.47

  5. Find Ksp. Use the ex key directly, or convert to base 10 first:

    Ksp=e−22.47=10−22.47/2.303=10−9.76=1.7×10−10\begin{aligned} K_{sp} &= e^{-22.47} \\ &= 10^{-22.47/2.303} = 10^{-9.76} \\ &= \boxed{1.7 \times 10^{-10}} \end{aligned}

  6. Compare. The OpenStax solubility-product table gives Ksp = 1.6 × 10⁻¹⁰ for AgCl, so the two data sets agree. ΔG° is positive, so Ksp ≪ 1 and AgCl is only slightly soluble.

Practice: at 25 °C, is ΔG° for dissolving this salt positive or negative?
  1. AgCl (Ksp = 1.6 × 10⁻¹⁰)
  2. NaCl (very soluble)
  3. PbI₂ (Ksp = 1.4 × 10⁻⁸)
  4. KCl (very soluble; the solution gets colder)
  5. CaCO₃ (Ksp = 8.7 × 10⁻⁹)
  6. ΔH° = +65 kJ/mol, ΔS° = +33 J/(K·mol)
  7. ΔH° = −20 kJ/mol, ΔS° = +10 J/(K·mol)
  8. ΔH° = +25 kJ/mol, ΔS° = +110 J/(K·mol)

Common mistakes

Mistake 1: “Endothermic dissolving can’t be spontaneous.” Spontaneity is decided by ΔG. With a large enough entropy increase, endothermic dissolving proceeds (NaCl, KCl, NH₄NO₃).

Mistake 2: “Entropy always increases when something dissolves.” Spreading out the ions increases entropy, but water molecules ordering around the ions decreases it. For small, highly charged ions the second effect can be large.

Mistake 3: “A salt with ΔG° > 0 doesn’t dissolve at all.” ΔG° > 0 only means K < 1. AgCl has Ksp = 1.6 × 10⁻¹⁰, and its saturated solution still contains 1.3 × 10⁻⁵ M Ag⁺.

Quick check

Dissolving NaCl is slightly endothermic, yet a lot of it dissolves at room temperature. What is the best explanation?

Dissolving an ionic compound is pictured in three steps. Which step releases energy?

A slightly soluble salt has Ksp = 2.0 × 10⁻⁷. ΔG° for dissolving it is:

At 25 °C, Ksp for AgBr is 5.0 × 10⁻¹³. ΔG° for dissolving it is about: (R = 8.314 J/(K·mol))

Dissolving a salt is endothermic (ΔH° > 0) with an entropy increase (ΔS° > 0). As the temperature rises, its solubility:

Which statement about the entropy change of dissolving is correct?

AP-style free response

AP-style free response

From OpenStax Appendix G: KCl(s) has ΔfH° = −436.5 kJ/mol and S° = 82.6 J/(K·mol); K⁺(aq) has ΔfH° = −252.4 kJ/mol and S° = 102.5 J/(K·mol); Cl⁻(aq) has ΔfH° = −167.2 kJ/mol and S° = 56.5 J/(K·mol).

(a) Calculate ΔH° at 25 °C for KCl(s) → K⁺(aq) + Cl⁻(aq). When KCl dissolves, does the temperature of the solution rise or fall?

(b) Calculate ΔS°, and explain its sign at the particle level.

(c) Calculate ΔG° at 25 °C, and decide whether dissolving is thermodynamically favorable.

(d) Explain why the result in (a) doesn’t contradict the conclusion in (c).

(e) Predict how the solubility of KCl changes as the temperature rises, and justify your prediction.

Model answer and scoring

(a) ΔH° = (−252.4) + (−167.2) − (−436.5) = +16.9 kJ/mol. Dissolving is endothermic, so the temperature of the solution falls.

(b) ΔS° = 102.5 + 56.5 − 82.6 = +76.4 J/(K·mol). The ions spread out from an orderly crystal into the solution, greatly increasing the number of possible arrangements, so entropy increases.

(c) ΔG° = 16.9 − 298 × 0.0764 = −5.9 kJ/mol (−6.0 from the free energies of formation); ΔG° < 0, so dissolving is thermodynamically favorable.

(d) Spontaneity is decided by ΔG, not by ΔH alone. The entropy increase makes −TΔS° (about −22.8 kJ/mol) more negative than ΔH° is positive, so the endothermic dissolving is still spontaneous.

(e) It increases. Dissolving is endothermic with ΔS° > 0, so heating makes ΔG° more negative and K larger (or: heating an endothermic process shifts the equilibrium toward dissolving).

Scoring: (a) ΔH° and the temperature change; (b) ΔS° with a particle-level explanation; (c) ΔG° with consistent units; (d) entropy-driven dissolving; (e) solubility rising with temperature for endothermic dissolving.

Summary

  • ΔH° of dissolving = separating the solute (endothermic) + separating the solvent (endothermic) + hydration (exothermic).
  • Dispersing the ions increases entropy; water ordering around the ions decreases it.
  • ΔG° = ΔH° − TΔS° and ΔG° = −RT ln Ksp: the more positive ΔG°, the smaller Ksp; endothermic dissolving can be driven by entropy.

AP tip: To explain why an endothermic process can be spontaneous, always mention entropy: ΔS > 0, and when T is large enough, TΔS > ΔH and ΔG < 0. When calculating ΔG° = ΔH° − TΔS°, convert ΔS° from J to kJ.

Practice this chapter: Unit 7 · Equilibrium · 12 questionsPractice this topic: U7.14 · Free Energy of Dissolution · 5 questions

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