Free Energy of Dissolution (College Board AP® Chemistry): Study Guide

Free energy of dissolution factors

The free energy of dissolution (ΔG°_diss)

• The free energy of dissolution, ΔG°_diss, shows whether dissolving a substance is thermodynamically favored

NaCl (s) → Na⁺ (aq) + Cl^– (aq)

• It is represented by the equation

ΔG°_diss = ΔH°_diss −TΔS°_diss

• ΔH°_diss = the standard enthalpy change of dissolution (kJ mol⁻¹)

• ΔS°_diss = the standard entropy change of dissolution (J K⁻¹ mol⁻¹)

• T = absolute temperature in kelvins (K)

• If the value for ΔG_diss is negative then dissolution is thermodynamically favorable

• If the value for ΔG_diss is positive then dissolution is not thermodynamically favorable

• The dissolution process is driven by

  • Enthalpic contributions

  • Entropic contributions

  • Or both of these

• A more negative ΔG°_diss means the salt dissolves to a greater extent

  • Predicting the sign is often hard, because the enthalpy and entropy terms of the three steps partly cancel

How does a solid dissolve?

There are three parts to a solid dissolving

Step 1: ΔH₁ and ΔS₁

Breaking the solid

• This step involves breaking the solid apart by overcoming the electrostatic forces between the ions

• This requires energy, which means that ΔH₁ is an endothermic process with a positive value

• The ions are freed from fixed lattice positions, so matter and energy become more dispersed (more microstates)

  • ΔS₁ is positive

Step 2: ΔH₂ and ΔS₂

Solvent preparing to dissolve the solid

• This step involves the solvent being prepared to dissolve the solid

• To make room for the solvent, the water molecules must move apart by overcoming the hydrogen bonds in water 

• This requires energy, which means that ΔH₂ is an endothermic process with a positive value

• The water molecules move apart, so they and their energy are more dispersed

  • ΔS₂ is positive 

Step 3: ΔH₃ and ΔS₃

Formation of ion–dipole forces between the ions and water molecules 

• The final step involves the formation of ion–dipole interactions between the water molecules and the ion

  • The δ− O atoms form an ion–dipole interaction with positive ions

  • The δ+ H atoms form an ion–dipole interaction with negative ions

  • Since attractive ion–dipole interactions form, ΔH₃ is exothermic (negative)

• In this step, the entropy change will actually be negative

  • This is because of the attraction between the water molecules and ions

  • So there is decreased freedom of movement and therefore number of microstates that can be formed

Enthalpy change, ΔH°_diss

• The enthalpy change during dissolution accounts for the heat absorbed or released in the process

  • If heat is absorbed, the dissolution is endothermic (ΔH_diss > 0)

  • If heat is released, the dissolution is exothermic (ΔH_diss < 0) 

• The enthalpy change for the dissolution of NaCl is positive, +4 kJ mol⁻¹, whereas the enthalpy change for the dissolution of MgCl₂ is −160 kJ mol⁻¹

  • This is due to the difference in the formation of the ion–dipole interaction

  • Given that Mg²⁺ is a smaller and more highly charged ion than Na⁺ more energy is released, so this process is more exothermic

Entropy change, ΔS°_diss

• The entropy change in each step affects whether dissolving is thermodynamically favored

  • Remember: entropy increases when matter and energy become more dispersed (more possible arrangements, or microstates)

• An increase in entropy (ΔS_diss > 0) means the particles and their energy are more dispersed

  • For many salts, such as NaCl, ΔS_diss is positive because ions held in a lattice become dispersed through the solution

  • For small, highly charged ions, such as Mg²⁺, ordering of water molecules around the ions can make ΔS_diss negative

• ΔS₁ and ΔS₂ are positive, and ΔS₃ is negative

  • For many salts, the first two terms outweigh ΔS₃, so ΔS_diss is positive

  • When ΔS₃ is larger, as for MgCl₂, ΔS_diss is negative

• Overall 

  • ΔH_diss = ΔH₁ + ΔH₂ + ΔH₃

  • ΔS_diss = ΔS₁ + ΔS₂ + ΔS₃

Dissolving a solid and entropy

When a solid such as NaCl dissolves to form a dilute solution, entropy usually increases because the solute particles become dispersed through a larger volume

Temperature and ΔG°_diss

• A dissolution that is not thermodynamically favorable at low temperature can become favorable at higher temperature if ΔS°_diss is positive

• If ΔS°_diss is positive, raising T makes −TΔS°_diss more negative, so ΔG°_diss becomes more negative

• For an endothermic dissolution with ΔS°_diss > 0, ΔG°_diss is negative only when TΔS°_diss is larger than ΔH°_diss

 Summary of factors affecting Gibbs free energy 

• Because the enthalpy and entropy terms can partly cancel, a calculation is needed to decide the sign of ΔG°_diss