Free Energy of Dissolution Factors
The Free Energy of Dissolution (ΔG°diss)
- The free energy of dissolution, (ΔGdiss) 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 ΔGdiss is negative then dissolution is thermodynamically favorable
- If the value for ΔGdiss 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 (ΔHdiss > 0)
- If heat is released, the dissolution is exothermic (ΔHdiss < 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 (ΔSdiss > 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 ΔSdiss is positive
- Overall
- ΔHdiss = ΔH1 + ΔH2 + ΔH3
- ΔSdiss = 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 TΔSdiss 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