Free Energy of Dissolution (College Board AP Chemistry)

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

ΔGdiss	=	ΔHdiss	−	TΔSdiss
–		+		+

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

 Summary of factors affecting Gibbs free energy 

If ΔH° ....	And if ΔS° ....	Then ΔG° is	Spontaneous?	Because
is negative < 0 exothermic	is positive > 0 more disorder	always negative < 0	Always	Forward reaction thermodynamically favorable at any T
is positive > 0 endothermic	is positive > 0 more disorder	negative at high T positive low T	Dependent on T	Thermodynamically favorable only at high T TΔS° > ΔH°

• 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