# Gibbs Free Energy and Thermodynamic Favorability

> AP Chemistry · Unit 9: Applications of Thermodynamics
> Source: https://www.owlsprep.com/study/ap-chemistry-u9-gibbs-free-energy-and-thermodynamic/

This module covers Gibbs free energy definition, the $\Delta G = \Delta H - T\Delta S$ relationship, calculation of standard $\Delta G^\circ$, the link between $\Delta G^\circ$ and equilibrium constant $K$, and rules for determining thermodynamic favorability for AP Chemistry.

**Prerequisites:** [Enthalpy and entropy change calculation from standard values](https://www.owlsprep.com/study/ap-chemistry-u6-enthalpy-change/); [Basic spontaneity concepts related to entropy](https://www.owlsprep.com/study/ap-chemistry-u9-entropy-entropy-change/); [Properties of reaction quotient Q and equilibrium constant K](https://www.owlsprep.com/study/ap-chemistry-u7-equilibrium-fundamentals/)

## Learning objectives

- Define Gibbs free energy and thermodynamic favorability for chemical processes
- Calculate ΔG using ΔG = ΔH - TΔS and determine thermodynamic favorability
- Calculate standard reaction ΔG° from standard Gibbs free energies of formation
- Relate ΔG, ΔG°, reaction quotient Q, and equilibrium constant K
- Calculate equilibrium constants from ΔG° and avoid common AP exam errors

## Core Concepts: Gibbs Free Energy and Favorability

Gibbs free energy ($G$) is a combined thermodynamic state function that links enthalpy ($\Delta H$) and entropy ($\Delta S$) to predict whether a process will be thermodynamically favorable for reactions at constant temperature and pressure — the condition that describes almost all reactions studied in AP Chemistry.

**Thermodynamically Favorable Process** — A process that will proceed to form products at given conditions without continuous input of external energy, once initiated. This is the AP Chemistry term for what many sources call a "spontaneous" process.

By universal convention: a negative $\Delta G$ corresponds to a thermodynamically favorable process, while a positive $\Delta G$ corresponds to an unfavorable process. When $\Delta G = 0$, the process is at equilibrium with no net change.

**Check your understanding**

Test your basic understanding:

1. If a process has $\Delta G = -15$ kJ/mol, what does this mean?

   - The process will occur very quickly
   - The process is thermodynamically favorable
   - The process is at equilibrium
   - The process will never occur

   *Answer:* The process is thermodynamically favorable

   *Why:* Correct! $\Delta G$ sign only indicates thermodynamic favorability, not reaction rate.

## The Fundamental Equation: $\Delta G = \Delta H - T\Delta S$

At constant temperature and pressure, the change in Gibbs free energy for any process is given by the equation:

$$\Delta G = \Delta H - T\Delta S$$

Where $\Delta G$ = change in Gibbs free energy (kJ/mol), $\Delta H$ = enthalpy change (kJ/mol), $T$ = absolute temperature in Kelvin, and $\Delta S$ = entropy change of the system.

- If $\Delta H < 0$ (exothermic) and $\Delta S > 0$: $\Delta G < 0$ at **all temperatures** (always favorable)
- If $\Delta H > 0$ (endothermic) and $\Delta S < 0$: $\Delta G > 0$ at **all temperatures** (always unfavorable)
- If $\Delta H$ and $\Delta S$ have matching signs: temperature determines the sign of $\Delta G$ and thus favorability

**Worked example:** For the evaporation of liquid ethanol to ethanol vapor at 1 atm, $\Delta H_{\text{vap}} = 38.6$ kJ/mol and $\Delta S_{\text{vap}} = 110$ J/mol·K. Is evaporation thermodynamically favorable at (a) 25°C, (b) 100°C?

1. Convert temperatures from Celsius to Kelvin:

   $$T_a = 25 + 273.15 = 298.15 \text{ K}; \quad T_b = 100 + 273.15 = 373.15 \text{ K}$$
2. Convert $\Delta S_{\text{vap}}$ to kJ/mol·K to match units of $\Delta H$:

   $$110 \text{ J/mol·K} = 0.110 \text{ kJ/mol·K}$$
3. Calculate $\Delta G$ at 25°C:

   $$\Delta G = 38.6 - (298.15)(0.110) = +5.8 \text{ kJ/mol}$$
4. Calculate $\Delta G$ at 100°C:

   $$\Delta G = 38.6 - (373.15)(0.110) = -2.4 \text{ kJ/mol}$$
5. Interpret: At 25°C, $\Delta G > 0$ so evaporation is unfavorable (liquid ethanol is stable); at 100°C, $\Delta G < 0$ so evaporation is favorable.

> **Exam tip:** Always convert $\Delta S$ from J/mol·K to kJ/mol·K before plugging into the equation. AP exam questions almost always give $\Delta S$ in joules and $\Delta H$ in kilojoules, so missing this conversion will give you a wrong sign and incorrect answer.

## Standard Gibbs Free Energy Calculations

Standard Gibbs free energy change ($\Delta G^\circ$) is the change in Gibbs free energy when reactants in their standard states (1 atm pressure, 1 M concentration, pure solid/liquid, 298 K by default) are converted to products in their standard states. There are two common methods to calculate $\Delta G^\circ$:

**Standard Gibbs Free Energy of Formation** — The $\Delta G$ for formation of 1 mole of a compound from its constituent elements in their standard states. By definition, $\Delta G^\circ_f = 0$ for any element in its standard state.

*Notation:* \Delta G^\circ_f

When using standard free energies of formation, the reaction $\Delta G^\circ$ is calculated as:

$$\Delta G^\circ_{\text{rxn}} = \sum n\Delta G^\circ_f(\text{products}) - \sum m\Delta G^\circ_f(\text{reactants})$$

**Worked example:** Calculate $\Delta G^\circ$ for the oxidation of iron (rust formation) at 298 K: $4\text{Fe}(s) + 3\text{O}_2(g) \rightarrow 2\text{Fe}_2\text{O}_3(s)$. Use the values: $\Delta G^\circ_f(\text{Fe}_2\text{O}_3(s)) = -742.2$ kJ/mol, $\Delta G^\circ_f(\text{Fe}(s)) = 0$ kJ/mol, $\Delta G^\circ_f(\text{O}_2(g)) = 0$ kJ/mol.

1. Write the $\Delta G^\circ_{\text{rxn}}$ formula for the balanced reaction:

   $$\Delta G^\circ_{\text{rxn}} = 2\Delta G^\circ_f(\text{Fe}_2\text{O}_3) - \left[4\Delta G^\circ_f(\text{Fe}) + 3\Delta G^\circ_f(\text{O}_2)\right]$$
2. Substitute the given values:

   $$\Delta G^\circ_{\text{rxn}} = 2(-742.2) - [4(0) + 3(0)]$$
3. Calculate the result:

   $$\Delta G^\circ_{\text{rxn}} = -1484.4 \text{ kJ/mol}$$
4. Interpret: The negative $\Delta G^\circ$ confirms that rust formation is thermodynamically favorable under standard conditions.

> **Exam tip:** Remember the minus sign in the $\Delta G^\circ_{\text{rxn}}$ formula applies to the entire sum of reactants. If any $\Delta G^\circ_f$ of reactants is negative, you will subtract a negative which equals adding that value — always write out signs explicitly to avoid arithmetic errors.

## $\Delta G$, $\Delta G^\circ$, and the Relationship to Equilibrium

$\Delta G^\circ$ only describes the Gibbs free energy change when the reaction is at standard state (all reactants and products at 1 M/1 atm, so $Q = 1$). For any non-standard conditions, we calculate $\Delta G$ using:

$$\Delta G = \Delta G^\circ + RT \ln Q$$

When a reaction reaches equilibrium, $\Delta G = 0$ (no net driving force) and $Q = K$ (the equilibrium constant). Substituting these values gives the key relationship connecting thermodynamics and equilibrium:

$$\Delta G^\circ = -RT \ln K$$

- $\Delta G^\circ < 0 \rightarrow \ln K > 0 \rightarrow K > 1$: Products are favored at equilibrium
- $\Delta G^\circ > 0 \rightarrow \ln K < 0 \rightarrow K < 1$: Reactants are favored at equilibrium
- $\Delta G^\circ = 0 \rightarrow \ln K = 0 \rightarrow K = 1$: Equal amounts of reactants and products at equilibrium

**Worked example:** For the dissolution of calcium hydroxide, $\text{Ca(OH)}_2(s) \rightleftharpoons \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq)$, $\Delta G^\circ = +39.0$ kJ/mol at 25°C. Calculate the solubility product constant $K_{\text{sp}}$ for calcium hydroxide at 25°C.

1. Convert temperature to Kelvin:

   $$T = 25 + 273.15 = 298.15 \text{ K}$$
2. Convert $\Delta G^\circ$ to J/mol to match units of $R$:

   $$39.0 \text{ kJ/mol} = 39000 \text{ J/mol}$$
3. Rearrange to solve for $\ln K_{\text{sp}}$:

   $$\ln K_{\text{sp}} = -\frac{\Delta G^\circ}{RT}$$
4. Substitute values ($R = 8.314$ J/mol·K):

   $$\ln K_{\text{sp}} = -\frac{39000}{(8.314)(298.15)} \approx -15.75$$
5. Exponentiate to get $K_{\text{sp}}$:

   $$K_{\text{sp}} = e^{-15.75} \approx 5.6 \times 10^{-7}$$

> **Exam tip:** Always use $R = 8.314$ J/mol·K for Gibbs free energy calculations, not $0.0821$ L·atm/mol·K (the gas constant used for ideal gas law problems). Using the wrong R will give you a K that is orders of magnitude off.

## Common pitfalls

- **Wrong:** Forgetting to convert ΔS from J/mol·K to kJ/mol·K in ΔG = ΔH - TΔS, leading to a ΔG with the wrong sign
  - Why it fails: AP questions almost always give ΔH in kJ and ΔS in J, so students plug in numbers without checking units
  - Correct: Always check units before plugging in; if they differ, convert ΔS to kJ to match ΔH
- **Wrong:** Claiming a reaction with ΔG > 0 will never occur at any observable rate
  - Why it fails: Students confuse thermodynamic favorability with kinetic feasibility. ΔG only describes favorability, not how fast the reaction proceeds
  - Correct: State the forward reaction is thermodynamically unfavorable, and note that this tells you nothing about the rate of the reaction
- **Wrong:** Claiming ΔG° = 0 at equilibrium
  - Why it fails: Students mix up the meaning of ΔG (any conditions) and ΔG° (only standard state)
  - Correct: Remember ΔG is always 0 at equilibrium; ΔG° is only 0 at equilibrium when K = 1
- **Wrong:** Treating any process with a positive ΔS of the system as always favorable
  - Why it fails: Students forget the second law refers to the entropy of the universe, not just the system
  - Correct: Always use the sign of ΔG, not just ΔS of the system, to determine thermodynamic favorability
- **Wrong:** Using R = 0.0821 L·atm/mol·K when calculating K from ΔG°
  - Why it fails: Students remember R from gas law problems and use it by mistake
  - Correct: Always reach for R = 8.314 J/mol·K for all Gibbs free energy calculations
- **Wrong:** Subtracting a negative ΔG°f value incorrectly, getting a positive ΔG° when it should be negative
  - Why it fails: Students forget the formula is products minus reactants, so a negative reactant ΔG°f becomes a positive term
  - Correct: Write all negative signs explicitly before plugging in numbers, e.g., ΔG° = products - (-50) = products + 50

## Cheatsheet

| Category | Formula | Key Notes |
| --- | --- | --- |
| Fundamental Gibbs free energy | \Delta G = \Delta H - T\Delta S | Convert ΔS to kJ/mol·K to match ΔH units |
| Favorability rule | ΔG < 0 = Favorable; ΔG = 0 = Equilibrium; ΔG > 0 = Unfavorable | Applies to all constant T,P processes |
| ΔG° from standard formation | \Delta G^\circ_{\text{rxn}} = \sum n\Delta G^\circ_f(\text{products}) - \sum m\Delta G^\circ_f(\text{reactants}) | ΔG°f = 0 for elements in standard state |
| ΔG for non-standard conditions | \Delta G = \Delta G^\circ + RT \ln Q | R = 8.314 J/mol·K, convert ΔG° to J |
| ΔG° and equilibrium constant | \Delta G^\circ = -RT \ln K | ΔG° < 0 → K > 1; ΔG° > 0 → K < 1 |
| Boundary temperature for favorability | T = \frac{\Delta H}{\Delta S} (at ΔG = 0) | ΔH-,ΔS- → favorable below T; ΔH+,ΔS+ → favorable above T |
| Coupled reactions | \Delta G_{\text{total}} = \sum \Delta G_{\text{individual}} | Gibbs free energy adds for sequential reactions |

## What's next

This topic forms the foundational link connecting thermodynamics to two core AP Chemistry topics: chemical equilibrium and electrochemistry, which are heavily tested on both MCQ and FRQ sections. Next, you will apply the relationship between ΔG° and K to predict how equilibrium constants change with temperature, a common multi-part FRQ skill. You will also connect ΔG to cell potential in electrochemistry, using the relation ΔG = -nFE to convert between cell voltage and Gibbs free energy change. Without mastering sign rules and unit conversions for Gibbs free energy, both of these topics will be far more difficult to solve correctly on the exam. This topic also completes the framework of thermodynamics started in Unit 6, giving a complete picture of energy and favorability for chemical processes.

- [Thermodynamic favorability versus rate](https://www.owlsprep.com/study/ap-chemistry-u9-thermodynamic-favorability-versus-rate/)
- [Free Energy and Equilibrium](https://www.owlsprep.com/study/ap-chemistry-u9-free-energy-and-equilibrium/)
- [Galvanic (voltaic) and electrolytic cells](https://www.owlsprep.com/study/ap-chemistry-u9-galvanic-and-electrolytic-cells/)

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