Free Energy of Dissolution (College Board AP® Chemistry)

Free Energy of Dissolution Factors

The Free Energy of Dissolution (ΔG°_diss)

• The free energy of dissolution, (ΔG_diss) is the thermodynamic parameter that characterizes the spontaneity of a dissolution process

NaCl (s) → Na⁺ (aq) + Cl^– (aq)

• It is represented by the equation

ΔG°_diss = ΔH°_diss −TΔS°_diss

• ΔH°_diss = the enthalpy change of…..

• ΔS°_diss = the entropy change of….

• T = temperature in Kelvin

• 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

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

• Since disorder is increasing, ΔS₁ will also be a positive value 

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

• Since disorder is increasing, ΔS₂ will be a positive value 

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 a ion-dipole interaction with positive ions

  • The δ+ H atoms form a ion-dipole interaction with negative ions

  • Since bonds are formed, ΔH₃ is an exothermic process with a negative value 

• 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

• As shown in the diagram above entropy plays a crucial role in dissolution

  • Remember entropy represents the degree of disorder in a system 

• An increase in entropy (ΔS_diss > 0) signifies a more disordered state

  • This fits in well with dissolution as the state change is from (s) to (aq) which represents an increase in disorder

NaCl (s) → Na⁺ (aq) + Cl^– (aq)

• First two terms in the diagram the positive values for ΔS₁ and ΔS₂ are both positive and ΔS₃ is generally negative

  • However, the magnitude of the first two terms (ΔS₁ and ΔS₂) outweighs ΔS₃ so the overall change for ΔS_diss is positive

• Overall 

  • ΔH_diss = ΔH₁ + ΔH₂ + ΔH₃

  • ΔS_diss = ΔS₁ + ΔS₂ + ΔS₃

Dissolving a solid and entropy

When a solid is dissolved in a solvent to form a dilute solution, the entropy increases as the particles become more disordered

Temperature, T, Effect on ΔG°_diss

• Temperature is a vital factor influencing ΔG°_diss

• While a process might be not thermodynamically favourable at low temperatures, an increase in temperature can make it thermodynamically favourable

• As T becomes larger, the term for TΔS_diss becomes more positive, so ΔG°_diss becomes more negative

• This temperature dependence is evident in industrial processes where controlling temperature becomes crucial for optimizing dissolution reactions

 Summary of factors affecting Gibbs free energy 

• As there are many factors that govern whether ΔG°_diss will be positive or negative it is necessary to carry out a calculation to show this

• For example, to calculate the ΔG°_diss for MgCl₂ at 298 K 

  • ΔH° = 160 kJ mol^-1

  • ΔS° = 114.7 J K^-1

• ΔG°_diss = ΔH° - TΔS°

• ΔG°_diss = 160 - 298 x  

  • Remember: divide by 1000 to convert J K^–1 to kJ K^–1

• ΔG°_diss = -125.8 kJ mol^-1

  • ΔG°_diss is a negative value so is thermodynamically favorable