# Free Energy of Dissolution

> AP Chemistry · Unit 7: Equilibrium
> Source: https://www.owlsprep.com/study/ap-chemistry-u7-free-energy-of-dissolution/

This guide covers Gibbs free energy change of dissolution ($\Delta G^\circ_{soln}$, $\Delta G_{soln}$), the relationship between $\Delta G^\circ_{soln}$ and $K_{sp}$, temperature-dependent solubility prediction, and classification of unsaturated/saturated/supersaturated solutions for AP Chemistry.

**Prerequisites:** [Gibbs free energy relation $\Delta G = \Delta G^\circ + RT \ln Q$](https://www.owlsprep.com/study/ap-chemistry-thermodynamics-gibbs-free-energy/); [Definition of the solubility product constant $K_{sp}$](https://www.owlsprep.com/study/ap-chemistry-u7-equilibrium-solubility-product/); Enthalpy and entropy changes for solution formation

## Learning objectives

- Relate standard free energy of dissolution to solubility product constant $K_{sp}$
- Calculate $\Delta G^\circ_{soln}$ from $\Delta H^\circ_{soln}$ and $\Delta S^\circ_{soln}$
- Predict how solubility changes with temperature
- Use non-standard $\Delta G_{soln}$ to classify solution saturation

## Core Definitions and Context

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.

## Standard Free Energy and the $\Delta G^\circ$-$K_{sp}$ Relationship

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.

**Worked example:** The $K_{sp}$ of lead(II) chloride ($PbCl_2$) at 298 K is $1.6 \times 10^{-5}$. Calculate $\Delta G^\circ_{soln}$ for $PbCl_2$ 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 $\ln K_{sp}$:

   $$\ln(1.6 \times 10^{-5}) = \ln(1.6) + \ln(10^{-5}) \approx 0.470 - 11.513 = -11.043$$
3. Substitute into the formula:

   $$\Delta G^\circ_{soln} = - (0.008314)(298)(-11.043) \approx +27.4 \text{ kJ/mol}$$
4. Interpret the result: $\Delta G^\circ_{soln}$ is positive, so dissolution is non-spontaneous under standard conditions, meaning $PbCl_2$ is sparingly soluble.

> **tip**
>
> When calculating $\Delta G^\circ_{soln}$ from $K_{sp}$, always confirm your units for $R$ match your desired final answer: use $R = 0.008314 \text{ kJ/(mol·K)}$ for $\Delta G$ in kJ/mol, which is what nearly all AP exam questions request.

## Calculating $\Delta G^\circ_{soln}$ from $\Delta H^\circ$ and $\Delta S^\circ$, Temperature Dependence

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.

**Worked example:** The dissolution of ammonium nitrate is $NH_4NO_3(s) \rightleftharpoons NH_4^+(aq) + NO_3^-(aq)$. At 298 K, $\Delta H^\circ_{soln} = +25.7 \text{ kJ/mol}$ and $\Delta S^\circ_{soln} = +108.7 \text{ J/(mol·K)}$. Calculate $\Delta G^\circ_{soln}$ at 298 K and 350 K, then predict how solubility changes with increasing temperature.

1. Convert $\Delta S^\circ_{soln}$ to kJ to match the units of $\Delta H^\circ$:

   $$\Delta S^\circ_{soln} = 108.7 / 1000 = 0.1087 \text{ kJ/(mol·K)}$$
2. Calculate $\Delta G^\circ$ at 298 K:

   $$\Delta G^\circ = 25.7 - (298)(0.1087) = 25.7 - 32.4 = -6.7 \text{ kJ/mol}$$
3. Calculate $\Delta G^\circ$ at 350 K:

   $$\Delta G^\circ = 25.7 - (350)(0.1087) = 25.7 - 38.0 = -12.3 \text{ kJ/mol}$$
4. Interpret the result: $\Delta G^\circ$ becomes more negative as temperature increases, so $K_{sp}$ increases, meaning solubility increases with increasing temperature.

> **tip**
>
> Always convert $\Delta S^\circ$ from J/(mol·K) to kJ/(mol·K) before substituting into $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$; a unit mismatch is the most common calculation error on this type of problem.

## Non-Standard Free Energy and Classifying Solution Saturation

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** ($Q > K_{sp}$)

**Worked example:** For $PbCl_2$ dissolution at 298 K, $\Delta G^\circ_{soln} = +27.4 \text{ kJ/mol}$. A solution has $[Pb^{2+}] = 0.001 \text{ M}$ and $[Cl^-] = 0.001 \text{ M}$. Calculate $\Delta G_{soln}$ 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 $\Delta G$ 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: $\Delta G_{soln}$ is negative, so dissolution is spontaneous, meaning the solution is unsaturated.

> **tip**
>
> If you forget what the sign of $\Delta G$ means for saturation, cross-check with $Q$: $Q < K_{sp}$ always means unsaturated, which always matches $\Delta G < 0$.

**Check your understanding**

1. A student measures $K_{sp}$ of silver sulfate ($Ag_2SO_4$) at 298 K as $1.2 \times 10^{-5}$. What is the correct $\Delta G^\circ_{soln}$ and solubility classification?

   - $\Delta G^\circ_{soln} = -28 \text{ kJ/mol}$, soluble
   - $\Delta G^\circ_{soln} = +28 \text{ kJ/mol}$, sparingly soluble
   - $\Delta G^\circ_{soln} = +2.8 \text{ kJ/mol}$, sparingly soluble
   - $\Delta G^\circ_{soln} = +28 \text{ kJ/mol}$, soluble

   *Answer:* $\Delta G^\circ_{soln} = +28 \text{ kJ/mol}$, sparingly soluble

   *Why:* Since $K_{sp} < 1$, $\ln K_{sp}$ is negative, so $\Delta G^\circ_{soln}$ is positive (+28 kJ/mol), which corresponds to sparingly soluble.

## Common pitfalls

- **Wrong:** Using $R = 0.0821 \text{ L·atm/(mol·K)}$ instead of $8.314 \text{ J/(mol·K)}$ when calculating $\Delta G$ from $K_{sp}$
  - Why it fails: 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
  - Correct: Always confirm the value of $R$ before starting: use $R = 8.314 \text{ J/(mol·K)}$ (or $0.008314 \text{ kJ/(mol·K)}$) for all Gibbs free energy calculations
- **Wrong:** Forgetting to convert $\Delta S^\circ$ from J/(mol·K) to kJ/(mol·K) when calculating $\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ$
  - Why it fails: $\Delta H^\circ$ 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
  - Correct: As a first step, divide $\Delta S^\circ$ by 1000 to convert to kJ/(mol·K) before substituting into the formula
- **Wrong:** Claiming a positive $\Delta G^\circ_{soln}$ means the compound never dissolves in water
  - Why it fails: 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
  - Correct: Interpret $\Delta G^\circ_{soln}$ as telling you the equilibrium extent of dissolution: positive $\Delta G^\circ$ means $K_{sp} < 1$, so only small amounts dissolve at equilibrium, not that no dissolution occurs
- **Wrong:** Getting the sign of $\Delta G^\circ_{soln}$ wrong when starting from $K_{sp}$ (e.g., getting a negative $\Delta G^\circ$ for $K_{sp} = 1.6 \times 10^{-5}$)
  - Why it fails: Students forget that $\ln$ of a number less than 1 is negative, so the two negative signs multiply to a positive $\Delta G^\circ$
  - Correct: After calculating, always check your sign: $K_{sp} < 1$ always gives positive $\Delta G^\circ_{soln}$, $K_{sp} > 1$ always gives negative $\Delta G^\circ_{soln}$
- **Wrong:** Claiming a positive $\Delta G_{soln}$ means the solution is unsaturated
  - Why it fails: Students mix up the direction of spontaneity: positive $\Delta G$ means the reverse reaction (precipitation) is spontaneous
  - Correct: Always pair $Q$ and $\Delta G$: $Q > K_{sp} \rightarrow \Delta G > 0 \rightarrow$ precipitation spontaneous $\rightarrow$ supersaturated
- **Wrong:** Using Celsius temperature instead of Kelvin in $\Delta G$ calculations
  - Why it fails: Students are used to Celsius for enthalpy problems where temperature differences are identical, but $\Delta G$ calculations require absolute temperature
  - Correct: Always convert any given Celsius temperature to Kelvin by adding 273 (acceptable for AP exams) before substituting into the formula

## Cheatsheet

| Category | Formula | Notes |
| --- | --- | --- |
| Standard $\Delta G^\circ$ from $K_{sp}$ | $\Delta G^\circ_{soln} = -RT \ln K_{sp}$ | $R = 0.008314 \text{ kJ/(mol·K)}$, $T$ in Kelvin. $\Delta G^\circ < 0 \rightarrow$ soluble, $\Delta G^\circ > 0 \rightarrow$ sparingly soluble |
| Standard $\Delta G^\circ$ from $\Delta H^\circ$ and $\Delta S^\circ$ | $\Delta G^\circ_{soln} = \Delta H^\circ_{soln} - T\Delta S^\circ_{soln}$ | Convert $\Delta S$ from J/(mol·K) to kJ/(mol·K) before calculation. $T$ in Kelvin |
| Non-standard $\Delta G$ for dissolution | $\Delta G_{soln} = \Delta G^\circ_{soln} + RT \ln Q$ | $Q =$ ion product = $[M^{y+}]^x[A^{x-}]^y$ for $M_xA_y(s)$ |
| Unsaturated solution | $\Delta G < 0$, $Q < K_{sp}$ | Forward dissolution is spontaneous; more solid will dissolve |
| Saturated solution | $\Delta G = 0$, $Q = K_{sp}$ | System at equilibrium; no net change in concentration |
| Supersaturated solution | $\Delta G > 0$, $Q > K_{sp}$ | Reverse precipitation/crystallization is spontaneous |
| Temperature dependence of solubility |  | $\Delta H^\circ_{soln} > 0 \rightarrow$ solubility increases with T; $\Delta H^\circ_{soln} < 0 \rightarrow$ 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 $K_{sp}$ to determine if a precipitate forms when two solutions are mixed. Without understanding how $\Delta G$ relates to $Q$ and $K_{sp}$ 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 $\Delta G^\circ$-$K$ relationship derived here.

- [Unit 7: Equilibrium Overview](https://www.owlsprep.com/study/ap-chemistry-u7-overview/)
- [Common Ion Effect](https://www.owlsprep.com/study/ap-chemistry-u7-common-ion-effect/)
- [pH and Solubility](https://www.owlsprep.com/study/ap-chemistry-u7-ph-and-solubility/)

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