Free Energy of Dissolution — AP Chemistry
1. Core Definitions and Context ★☆☆☆☆ ⏱ 2 min
Free energy of dissolution refers to the Gibbs free energy change that occurs when 1 mole of solute dissolves in a solvent at constant temperature and pressure. For AP Chemistry, standard state conditions are 1 atm pressure, 1 M solute concentration, and 298 K unless stated otherwise. The standard free energy of dissolution is denoted $\Delta G^\circ_{soln}$, while non-standard free energy change at non-standard concentrations is called $\Delta G_{soln}$.
This topic accounts for 7–9% of the total AP Chemistry exam weight, appearing in both multiple-choice and free-response sections. It bridges core thermodynamics and equilibrium, two central big ideas of the AP Chemistry curriculum.
2. Standard Free Energy and the $\Delta G^\circ$-$K_{sp}$ Relationship ★★☆☆☆ ⏱ 4 min
Dissolution is a homogeneous equilibrium process, so the general relationship between standard Gibbs free energy change and the equilibrium constant applies directly. For the dissolution of a generic ionic solid $M_xA_y(s)$:
M_xA_y(s) \rightleftharpoons xM^{y+}(aq) + yA^{x-}(aq)
The equilibrium constant for this reaction is the solubility product constant $K_{sp}$. The core formula relating $\Delta G^\circ_{soln}$ to $K_{sp}$ is:
\Delta G^\circ_{soln} = -RT \ln K_{sp}
Where $R = 8.314 \text{ J/(mol·K)} = 0.008314 \text{ kJ/(mol·K)}$, and $T$ is absolute temperature in Kelvin. This formula directly tells us solubility under standard conditions:
- If $\Delta G^\circ_{soln} < 0$, dissolution is spontaneous, so $K_{sp} > 1$, and the compound is soluble.
- If $\Delta G^\circ_{soln} > 0$, dissolution is non-spontaneous under standard conditions, so $K_{sp} < 1$, and the compound is sparingly soluble or insoluble.
- At $\Delta G^\circ_{soln} = 0$, $K_{sp} = 1$, the boundary between soluble and insoluble behavior.
3. Calculating $\Delta G^\circ_{soln}$ from $\Delta H^\circ$ and $\Delta S^\circ$, Temperature Dependence ★★★☆☆ ⏱ 5 min
When $K_{sp}$ is not provided, you can calculate $\Delta G^\circ_{soln}$ using the fundamental Gibbs free energy relation:
\Delta G^\circ_{soln} = \Delta H^\circ_{soln} - T\Delta S^\circ_{soln}
Here, $\Delta H^\circ_{soln}$ is the standard enthalpy change when 1 mole of solute dissolves, and $\Delta S^\circ_{soln}$ is the standard entropy change of dissolution. This formula lets you quantitatively predict how solubility changes with temperature, a common AP exam question:
- If $\Delta H^\circ_{soln}$ is positive (endothermic dissolution), increasing $T$ makes $\Delta G^\circ_{soln}$ more negative, $K_{sp}$ increases, and solubility increases.
- If $\Delta H^\circ_{soln}$ is negative (exothermic dissolution), increasing $T$ makes $\Delta G^\circ_{soln}$ more positive, $K_{sp}$ decreases, and solubility decreases.
This matches Le Chatelier’s principle but gives a quantitative framework for calculations.
4. Non-Standard Free Energy and Classifying Solution Saturation ★★★☆☆ ⏱ 5 min
When a solution is not at equilibrium (not saturated), we use the non-standard Gibbs free energy change of dissolution, which follows the general relation:
\Delta G_{soln} = \Delta G^\circ_{soln} + RT \ln Q
Where $Q$ is the reaction quotient, calculated the same way as $K_{sp}$ but using current ion concentrations instead of equilibrium concentrations. The sign of $\Delta G_{soln}$ tells us which direction the reaction proceeds to reach equilibrium:
- $\Delta G_{soln} < 0$: Dissolution is spontaneous, so more solid will dissolve → solution is **unsaturated** ($Q < K_{sp}$)
- $\Delta G_{soln} = 0$: System is at equilibrium, no net change → solution is **saturated** ($Q = K_{sp}$)
- $\Delta G_{soln} > 0$: Precipitation (reverse reaction) is spontaneous, so ions will precipitate → solution is **supersaturated** ($Q > K_{sp}$)
Common Pitfalls
Why: Students remember $R$ from gas law problems and use it by mistake, leading to a $\Delta G$ value that is 3 orders of magnitude incorrect
Why: $\Delta H^\circ$ is almost always reported in kJ/mol, while $\Delta S^\circ$ is reported in J/(mol·K), so leaving $\Delta S$ in J gives a $\Delta G$ value 1000x too large
Why: Students confuse standard $\Delta G^\circ$ with non-standard $\Delta G$; many sparingly soluble compounds dissolve to a small extent even if $\Delta G^\circ$ is positive
Why: Students forget that $\ln$ of a number less than 1 is negative, so the two negative signs multiply to a positive $\Delta G^\circ$
Why: Students mix up the direction of spontaneity: positive $\Delta G$ means the reverse reaction (precipitation) is spontaneous
Why: Students are used to Celsius for enthalpy problems where temperature differences are identical, but $\Delta G$ calculations require absolute temperature