Free Energy of Dissolution

AP Chemistry· AP Chemistry CED — Equilibrium· 14 min read

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 $Δ G_{so l n}^{\circ}$ , while non-standard free energy change at non-standard concentrations is called $Δ G_{so l n}$ .

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 $M_{x} A_{y} (s)$ :

M_xA_y(s) \rightleftharpoons xM^{y+}(aq) + yA^{x-}(aq)

The equilibrium constant for this reaction is the solubility product constant $K_{s p}$ . The core formula relating $Δ G_{so l n}^{\circ}$ to $K_{s p}$ is:

\Delta G^\circ_{soln} = -RT \ln K_{sp}

Where $R = 8.314 J/(mol \cdotp K) = 0.008314 kJ/(mol \cdotp K)$ , and $T$ is absolute temperature in Kelvin. This formula directly tells us solubility under standard conditions:

If $Δ G_{so l n}^{\circ} < 0$ , dissolution is spontaneous, so $K_{s p} > 1$ , and the compound is soluble.
If $Δ G_{so l n}^{\circ} > 0$ , dissolution is non-spontaneous under standard conditions, so $K_{s p} < 1$ , and the compound is sparingly soluble or insoluble.
At $Δ G_{so l n}^{\circ} = 0$ , $K_{s p} = 1$ , the boundary between soluble and insoluble behavior.

📐 Worked Example

The $K_{s p}$ of lead(II) chloride ( $P b C l_{2}$ ) at 298 K is $1.6 \times 1 0^{- 5}$ . Calculate $Δ G_{so l n}^{\circ}$ for $P b C l_{2}$ at this temperature, and classify it as soluble or insoluble under standard conditions.

1
List given values:
$K_{sp} = 1.6 \times 10^{-5}, T = 298 \text{ K}, R = 0.008314 \text{ kJ/(mol·K)}$
2
Calculate $ln K_{s p}$ :
$\ln(1.6 \times 10^{-5}) = \ln(1.6) + \ln(10^{-5}) \approx 0.470 - 11.513 = -11.043$
3
Substitute into the formula:
$\Delta G^\circ_{soln} = - (0.008314)(298)(-11.043) \approx +27.4 \text{ kJ/mol}$
4
Interpret the result: $Δ G_{so l n}^{\circ}$ is positive, so dissolution is non-spontaneous under standard conditions, meaning $P b C l_{2}$ is sparingly soluble.

3. Calculating $\Delta G^\circ_{soln}$ from $\Delta H^\circ$ and $\Delta S^\circ$, Temperature Dependence★★★☆☆⏱ 5 min

When $K_{s p}$ is not provided, you can calculate $Δ G_{so l n}^{\circ}$ using the fundamental Gibbs free energy relation:

\Delta G^\circ_{soln} = \Delta H^\circ_{soln} - T\Delta S^\circ_{soln}

Here, $Δ H_{so l n}^{\circ}$ is the standard enthalpy change when 1 mole of solute dissolves, and $Δ S_{so l n}^{\circ}$ is the standard entropy change of dissolution. This formula lets you quantitatively predict how solubility changes with temperature, a common AP exam question:

If $Δ H_{so l n}^{\circ}$ is positive (endothermic dissolution), increasing $T$ makes $Δ G_{so l n}^{\circ}$ more negative, $K_{s p}$ increases, and solubility increases.
If $Δ H_{so l n}^{\circ}$ is negative (exothermic dissolution), increasing $T$ makes $Δ G_{so l n}^{\circ}$ more positive, $K_{s p}$ decreases, and solubility decreases.

This matches Le Chatelier’s principle but gives a quantitative framework for calculations.

📐 Worked Example

The dissolution of ammonium nitrate is $N H_{4} N O_{3} (s) ⇌ N H_{4}^{+} (a q) + N O_{3}^{-} (a q)$ . At 298 K, $Δ H_{so l n}^{\circ} = + 25.7 kJ/mol$ and $Δ S_{so l n}^{\circ} = + 108.7 J/(mol \cdotp K)$ . Calculate $Δ G_{so l n}^{\circ}$ at 298 K and 350 K, then predict how solubility changes with increasing temperature.

1
Convert $Δ S_{so l n}^{\circ}$ to kJ to match the units of $Δ H^{\circ}$ :
$\Delta S^\circ_{soln} = 108.7 / 1000 = 0.1087 \text{ kJ/(mol·K)}$
2
Calculate $Δ G^{\circ}$ at 298 K:
$\Delta G^\circ = 25.7 - (298)(0.1087) = 25.7 - 32.4 = -6.7 \text{ kJ/mol}$
3
Calculate $Δ G^{\circ}$ at 350 K:
$\Delta G^\circ = 25.7 - (350)(0.1087) = 25.7 - 38.0 = -12.3 \text{ kJ/mol}$
4
Interpret the result: $Δ G^{\circ}$ becomes more negative as temperature increases, so $K_{s p}$ increases, meaning solubility increases with increasing temperature.

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 $K_{s p}$ but using current ion concentrations instead of equilibrium concentrations. The sign of $Δ G_{so l n}$ tells us which direction the reaction proceeds to reach equilibrium:

$Δ G_{so l n} < 0$ : Dissolution is spontaneous, so more solid will dissolve → solution is unsaturated ( $Q < K_{s p}$ )
$Δ G_{so l n} = 0$ : System is at equilibrium, no net change → solution is saturated ( $Q = K_{s p}$ )
$Δ G_{so l n} > 0$ : Precipitation (reverse reaction) is spontaneous, so ions will precipitate → solution is supersaturated ( $Q > K_{s p}$ )

📐 Worked Example

For $P b C l_{2}$ dissolution at 298 K, $Δ G_{so l n}^{\circ} = + 27.4 kJ/mol$ . A solution has $[P b^{2 +}] = 0.001 M$ and $[C l^{-}] = 0.001 M$ . Calculate $Δ G_{so l n}$ and classify the solution.

1
Write the expression for $Q$ :
$Q = [Pb^{2+}][Cl^-]^2$
2
Calculate $Q$ :
$Q = (0.001)(0.001)^2 = 1 \times 10^{-9}$
3
Substitute into the $Δ G$ formula:
$\Delta G_{soln} = 27.4 + (0.008314)(298)(\ln 1 \times 10^{-9}) \approx 27.4 + (2.478)(-20.72) \approx -23.9 \text{ kJ/mol}$
4
Interpret: $Δ G_{so l n}$ is negative, so dissolution is spontaneous, meaning the solution is unsaturated.

✓ Quick check

A student measures $K_{s p}$ of silver sulfate ( $A g_{2} S O_{4}$ ) at 298 K as $1.2 \times 1 0^{- 5}$ . What is the correct $Δ G_{so l n}^{\circ}$ and solubility classification?
- $Δ G_{so l n}^{\circ} = - 28 kJ/mol$ , soluble
- $Δ G_{so l n}^{\circ} = + 28 kJ/mol$ , sparingly soluble
- $Δ G_{so l n}^{\circ} = + 2.8 kJ/mol$ , sparingly soluble
- $Δ G_{so l n}^{\circ} = + 28 kJ/mol$ , soluble
Reveal answer
1 —
Since $K_{s p} < 1$ , $ln K_{s p}$ is negative, so $Δ G_{so l n}^{\circ}$ is positive (+28 kJ/mol), which corresponds to sparingly soluble.

5. Common Pitfalls

Wrong move:

Using $R = 0.0821 L \cdotp atm/(mol \cdotp K)$ instead of $8.314 J/(mol \cdotp K)$ when calculating $Δ G$ from $K_{s p}$

Why:

Students remember $R$ from gas law problems and use it by mistake, leading to a $Δ G$ value that is 3 orders of magnitude incorrect

Correct move:

Always confirm the value of $R$ before starting: use $R = 8.314 J/(mol \cdotp K)$ (or $0.008314 kJ/(mol \cdotp K)$ ) for all Gibbs free energy calculations

Wrong move:

Forgetting to convert $Δ S^{\circ}$ from J/(mol·K) to kJ/(mol·K) when calculating $Δ G^{\circ} = Δ H^{\circ} - T Δ S^{\circ}$

Why:

$Δ H^{\circ}$ is almost always reported in kJ/mol, while $Δ S^{\circ}$ is reported in J/(mol·K), so leaving $Δ S$ in J gives a $Δ G$ value 1000x too large

Correct move:

As a first step, divide $Δ S^{\circ}$ by 1000 to convert to kJ/(mol·K) before substituting into the formula

Wrong move:

Claiming a positive $Δ G_{so l n}^{\circ}$ means the compound never dissolves in water

Why:

Students confuse standard $Δ G^{\circ}$ with non-standard $Δ G$ ; many sparingly soluble compounds dissolve to a small extent even if $Δ G^{\circ}$ is positive

Correct move:

Interpret $Δ G_{so l n}^{\circ}$ as telling you the equilibrium extent of dissolution: positive $Δ G^{\circ}$ means $K_{s p} < 1$ , so only small amounts dissolve at equilibrium, not that no dissolution occurs

Wrong move:

Getting the sign of $Δ G_{so l n}^{\circ}$ wrong when starting from $K_{s p}$ (e.g., getting a negative $Δ G^{\circ}$ for $K_{s p} = 1.6 \times 1 0^{- 5}$ )

Why:

Students forget that $ln$ of a number less than 1 is negative, so the two negative signs multiply to a positive $Δ G^{\circ}$

Correct move:

After calculating, always check your sign: $K_{s p} < 1$ always gives positive $Δ G_{so l n}^{\circ}$ , $K_{s p} > 1$ always gives negative $Δ G_{so l n}^{\circ}$

Wrong move:

Claiming a positive $Δ G_{so l n}$ means the solution is unsaturated

Why:

Students mix up the direction of spontaneity: positive $Δ G$ means the reverse reaction (precipitation) is spontaneous

Correct move:

Always pair $Q$ and $Δ G$ : $Q > K_{s p} \to Δ G > 0 \to$ precipitation spontaneous $\to$ supersaturated

Wrong move:

Using Celsius temperature instead of Kelvin in $Δ G$ calculations

Why:

Students are used to Celsius for enthalpy problems where temperature differences are identical, but $Δ G$ calculations require absolute temperature

Correct move:

Always convert any given Celsius temperature to Kelvin by adding 273 (acceptable for AP exams) before substituting into the formula

6. Quick Reference Cheatsheet

Category	Formula	Notes
Standard $Δ G^{\circ}$ from $K_{s p}$	$Δ G_{so l n}^{\circ} = - R T ln K_{s p}$	$R = 0.008314 kJ/(mol \cdotp K)$ , $T$ in Kelvin. $Δ G^{\circ} < 0 \to$ soluble, $Δ G^{\circ} > 0 \to$ sparingly soluble
Standard $Δ G^{\circ}$ from $Δ H^{\circ}$ and $Δ S^{\circ}$	$Δ G_{so l n}^{\circ} = Δ H_{so l n}^{\circ} - T Δ S_{so l n}^{\circ}$	Convert $Δ S$ from J/(mol·K) to kJ/(mol·K) before calculation. $T$ in Kelvin
Non-standard $Δ G$ for dissolution	$Δ G_{so l n} = Δ G_{so l n}^{\circ} + R T ln Q$	$Q =$ ion product = $[M^{y +}]^{x} [A^{x -}]^{y}$ for $M_{x} A_{y} (s)$
Unsaturated solution	$Δ G < 0$ , $Q < K_{s p}$	Forward dissolution is spontaneous; more solid will dissolve
Saturated solution	$Δ G = 0$ , $Q = K_{s p}$	System at equilibrium; no net change in concentration
Supersaturated solution	$Δ G > 0$ , $Q > K_{s p}$	Reverse precipitation/crystallization is spontaneous
Temperature dependence of solubility		$Δ H_{so l n}^{\circ} > 0 \to$ solubility increases with T; $Δ H_{so l n}^{\circ} < 0 \to$ solubility decreases with T

What's Next

Free energy of dissolution bridges core thermodynamics and solubility equilibrium, and it is a prerequisite for multiple key topics that follow in AP Chemistry. Immediately after mastering this topic, you will move on to predicting precipitation reactions, where you compare $Q$ and $K_{s p}$ to determine if a precipitate forms when two solutions are mixed. Without understanding how $Δ G$ relates to $Q$ and $K_{s p}$ for dissolution, you cannot correctly predict precipitation behavior, a common AP FRQ topic. This topic also feeds into the broader study of colligative properties, where free energy changes of dissolution drive boiling point elevation and freezing point depression, and into acid-base equilibrium, where the dissociation of weak acids and bases follows the same $Δ G^{\circ}$ - $K$ relationship derived here.