---
title: "AP Chemistry 9.5: Free Energy and Equilibrium"
description: "Review AP Chemistry 9.5 free energy and equilibrium, including thermodynamically favored processes, K, ΔG°, ΔG° = -RT ln K, K = e^(-ΔG°/RT), and qualitative estimates."
canonical: "https://fiveable.me/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr"
type: "study-guide"
subject: "AP Chemistry"
unit: "Unit 9 – Thermodynamics and Electrochemistry"
lastUpdated: "2026-06-09"
---

# AP Chemistry 9.5: Free Energy and Equilibrium

## Summary

Review AP Chemistry 9.5 free energy and equilibrium, including thermodynamically favored processes, K, ΔG°, ΔG° = -RT ln K, K = e^(-ΔG°/RT), and qualitative estimates.

## Guide

Free energy and equilibrium are two ways of describing whether a reaction favors [products](/ap-chem/key-terms/products "fv-autolink") or reactants. When the standard free energy change, $\Delta G^\circ$, is negative, products are favored and $K$ is greater than 1; when $\Delta G^\circ$ is positive, reactants are favored and $K$ is less than 1. For [AP Chemistry](/ap-chem "fv-autolink"), use both the sign of $\Delta G^\circ$ and the size of $K$ to support the same conclusion.

## Why This Matters for the AP Chemistry Exam

This topic ties Gibbs free energy from earlier in [Unit 9](/ap-chem/unit-9 "fv-autolink") back to the [equilibrium](/ap-chem/unit-7/reaction-quotient-le-chateliers-principle/study-guide/JFx1InPfZCZ9SugPKDCE "fv-autolink") ideas from Unit 7. On the AP Chemistry exam, you may need to justify whether a process is thermodynamically favored using the relationships between K, ΔG°, and T, and explain your reasoning with chemical principles or math. Many questions ask you to reason about the size and sign of K from ΔG° (or the reverse) instead of just plugging into a formula, so being able to estimate K qualitatively is just as important as the calculation.

## Key Takeaways

- "Thermodynamically favored" (ΔG° < 0) means products are favored at equilibrium, so K > 1 under [standard conditions](/ap-chem/key-terms/standard-conditions "fv-autolink").
- The core equations are [ΔG° = -RT ln K](/ap-chem/key-terms/g-rt-ln-k "fv-autolink") and K = e^(-ΔG°/RT), with R = 8.314 J·mol⁻¹·K⁻¹ and T in Kelvin.
- A negative ΔG° gives K > 1 (products favored); a positive ΔG° gives K < 1 (reactants favored).
- When ΔG° is near zero, K is close to 1. When ΔG° is much larger or much smaller than RT, K deviates strongly from 1.
- ΔG (no naught) describes free energy at any conditions; ΔG° describes the standard-state case. They connect through Q: ΔG = ΔG° + RT ln Q.
- At equilibrium, ΔG = 0 and Q = K, which is exactly how the ΔG°-K equation is derived.

## Two Ways to Define Equilibrium

Unit 7 used a **kinetic definition** of equilibrium: the point where the forward and reverse reactions happen at the same rate, so the [concentrations](/ap-chem/unit-3/beer-lambert-law/study-guide/smCHzraorVz6qlWW1oeB "fv-autolink") of products and reactants stop changing. Equilibrium does not mean nothing is happening. Both reactions are still going, they just cancel out.

There is also a **thermodynamic definition**: equilibrium is the point of minimum free energy. While a reaction moves toward equilibrium on its own, ΔG (no naught symbol) is less than zero, so the reaction releases free energy. Once it reaches [equilibrium concentrations](/ap-chem/unit-7/magnitude-equilibrium-constant/study-guide/dvXT7PLceyYd2QH8KiV4 "fv-autolink"), moving further in either direction would require ΔG to be positive, meaning external energy would be needed.

Picture a graph with free energy (G) on the y-axis and the extent of reaction on the x-axis, running from 100% reactants to 100% products.

A few things to keep straight when reading that kind of graph:

- ΔG is not the height of the curve. It is the slope (the rate of change). When ΔG < 0, the curve is going downhill; when ΔG > 0, it is going uphill.
- The far left point is 100% reactants and the far right is 100% products. The difference in height between those two ends reflects the sign of ΔG° for the reaction, calculated as ΔG° = ΣnΔG°f (products) - ΣnΔG°f (reactants).
- The lowest point on the curve is the equilibrium point, where ΔG = 0.

For a reaction where products sit lower than reactants, ΔG° is negative (thermodynamically favored). The [system](/ap-chem/key-terms/system "fv-autolink") slides downhill toward the minimum, making products, until it hits the equilibrium point. Past that point ΔG > 0, so pushing the reaction further toward products would require adding energy.

For a reaction where products sit higher than reactants, ΔG° is positive (not favored). The equilibrium point lands closer to the [reactant](/ap-chem/unit-7/representations-equilibrium/study-guide/wLQChBkGSKiEP5xvlXB8 "fv-autolink") side, so there are many more reactants than products at equilibrium. A reaction that is not favored ends up reactant-heavy at equilibrium.

## Relationship Between ΔG°, ΔG, and K

ΔG measures free energy change at **nonstandard** conditions, so you connect ΔG° and ΔG through Q, the reaction quotient:

ΔG = ΔG° + RT ln Q

To find Q, plug non-equilibrium concentrations or pressures into the [equilibrium expression](/ap-chem/key-terms/equilibrium-expression "fv-autolink") (the [law of mass action](/ap-chem/key-terms/law-of-mass-action "fv-autolink")). R is the gas constant (8.314 J·mol⁻¹·K⁻¹), and T is the temperature in Kelvin.

To link ΔG° directly to K, use the two conditions that are true at equilibrium: ΔG = 0 and Q = K. Substituting in:

0 = ΔG° + RT ln K

**ΔG° = -RT ln K**

Solving for K:

-ΔG° = RT ln K

-ΔG°/RT = ln K

**K = e^(-ΔG°/RT)**

These equations show a direct relationship between ΔG° and K:

- A negative ΔG° makes -ΔG°/RT positive, so K = e^(positive number), which is greater than 1 (products favored).
- A positive ΔG° makes -ΔG°/RT negative, so K = e^(negative number), which is less than 1 (reactants favored).
- The more negative ΔG° gets, the larger K becomes; the more positive ΔG° gets, the smaller K becomes.

## Estimating K from ΔG° Without a Calculator

A lot of questions want reasoning, not just a number. The size of K compared to 1 depends on how ΔG° compares to RT:

- ΔG° near zero means K is close to 1, so products and reactants are present in similar amounts at equilibrium.
- ΔG° much smaller than RT (a large negative value) means K is much greater than 1, so products are strongly favored.
- ΔG° much larger than RT (a large positive value) means K is much less than 1, so reactants strongly outweigh products.

Being able to say "ΔG° is large and negative, so K is very large" without computing the exact value is the kind of qualitative reasoning the exam rewards.

## How to Use This on the AP Chemistry Exam

### Problem Solving

- Watch your units. ΔG° often comes in kJ·mol⁻¹ while R is in J·mol⁻¹·K⁻¹. Convert kJ to J before using ΔG° = -RT ln K.
- Always use temperature in Kelvin.
- You can find ΔG° first from standard free energies of formation (ΔG° = ΣnΔG°f products - ΣnΔG°f reactants), then plug it into the K equation.

### Free Response

- When asked to justify whether a process is favored, connect the sign of ΔG° to the value of K and to which side is favored at equilibrium. Use chemical reasoning or the math relationship, not just a restated definition.
- If you only need to know whether K is greater or less than 1, you do not have to calculate it. The sign of ΔG° is enough.

### Common Trap

- Q and K are not the same. Q can be any value at any moment; K is the specific value of Q at equilibrium. ΔG uses Q, while ΔG° uses K.

## Common Misconceptions

- Equilibrium does not mean the reaction stopped. The forward and reverse reactions are still running at equal rates.
- ΔG and ΔG° are different. ΔG° applies only to standard conditions; ΔG applies to whatever conditions the system is actually in. Mixing them up breaks the math.
- A negative ΔG° does not mean the reaction goes to completion. It means K > 1, so products are favored, but there is still some reactant left at equilibrium.
- A positive ΔG° does not mean no reaction happens at all. It means K < 1, so the reaction still proceeds a little, just reactant-favored.
- ΔG° tells you the position of equilibrium, not the speed. A favored reaction can still be extremely slow if it is under kinetic control.
- On the free energy vs. extent of reaction graph, ΔG is the slope of the curve, not the height. The lowest point is equilibrium, where the slope (ΔG) is zero.

## Related AP Chemistry Guides

- [9.1 Introduction to Entropy](/ap-chem/unit-9/intro-entropy/study-guide/rrwnj8YrrJ2xOtgibBsL)
- [9.3 Gibbs Free Energy and Thermodynamic Favorability](/ap-chem/unit-9/gibbs-free-energy-thermodynamic-favorability/study-guide/hCJVI2XJaSGmj1c3zvrO)
- [9.2 Absolute Entropy and Entropy Change](/ap-chem/unit-9/absolute-entropy-entropy-change/study-guide/RhcmO6DuTuEoGzcCk2zG)
- [9.8 Galvanic (Voltaic) and Electrolytic Cells](/ap-chem/unit-9/galvanic-voltaic-electrolyticlls/study-guide/egSkWaC0jSmJvCszUAkK)
- [9.4 Thermodynamic and Kinetic Control](/ap-chem/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA)
- [Unit 9 Review: Thermodynamics and Electrochemistry](/ap-chem/unit-9/review/study-guide/DxMXST1Cs4yxSBPsuiCN)

## Vocabulary

- **RT (gas constant × temperature)**: The product of the universal gas constant and absolute temperature; used in the relationship between K and ΔG°.
- **equilibrium constant**: A numerical value that expresses the ratio of products to reactants at equilibrium, indicating the extent to which a reaction proceeds.
- **standard Gibbs free energy change**: The change in free energy under standard conditions; negative values indicate thermodynamically favored processes that favor products.
- **standard conditions**: The reference conditions (typically 25°C, 1 M concentration, 1 atm pressure) under which ΔG° and K are evaluated.
- **thermodynamically favored**: A reaction or process that has a negative Gibbs free energy (ΔG < 0) and is spontaneous under given conditions.

## FAQs

### How are ΔG° and K related in AP Chemistry?

Standard free energy and the equilibrium constant are related by ΔG° = -RT ln K. The equation connects thermodynamic favorability to whether products or reactants are favored at equilibrium.

### What does a negative ΔG° mean for K?

A negative ΔG° means K is greater than 1, so products are favored at equilibrium under standard conditions.

### What does a positive ΔG° mean for K?

A positive ΔG° means K is less than 1, so reactants are favored at equilibrium under standard conditions.

### What does it mean if ΔG° is close to zero?

If ΔG° is close to zero, K is close to 1. That means products and reactants are present in more comparable amounts at equilibrium.

### What is the difference between ΔG and ΔG°?

ΔG describes free energy change under the actual conditions of the system, while ΔG° describes standard-state conditions. They connect through ΔG = ΔG° + RT ln Q.

### How is AP Chemistry 9.5 tested?

AP Chemistry 9.5 is tested through sign and size reasoning with ΔG° and K, calculations using ΔG° = -RT ln K, unit conversions, and explanations of which side is favored at equilibrium.

## Structured Data

```json
{"@context":"https://schema.org","@type":"FAQPage","inLanguage":"en","mainEntity":[{"@type":"Question","@id":"https://fiveable.me/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr#how-are-g-and-k-related-in-ap-chemistry","name":"How are ΔG° and K related in AP Chemistry?","acceptedAnswer":{"@type":"Answer","text":"Standard free energy and the equilibrium constant are related by ΔG° = -RT ln K. The equation connects thermodynamic favorability to whether products or reactants are favored at equilibrium."}},{"@type":"Question","@id":"https://fiveable.me/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr#what-does-a-negative-g-mean-for-k","name":"What does a negative ΔG° mean for K?","acceptedAnswer":{"@type":"Answer","text":"A negative ΔG° means K is greater than 1, so products are favored at equilibrium under standard conditions."}},{"@type":"Question","@id":"https://fiveable.me/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr#what-does-a-positive-g-mean-for-k","name":"What does a positive ΔG° mean for K?","acceptedAnswer":{"@type":"Answer","text":"A positive ΔG° means K is less than 1, so reactants are favored at equilibrium under standard conditions."}},{"@type":"Question","@id":"https://fiveable.me/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr#what-does-it-mean-if-g-is-close-to-zero","name":"What does it mean if ΔG° is close to zero?","acceptedAnswer":{"@type":"Answer","text":"If ΔG° is close to zero, K is close to 1. That means products and reactants are present in more comparable amounts at equilibrium."}},{"@type":"Question","@id":"https://fiveable.me/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr#what-is-the-difference-between-g-and-g","name":"What is the difference between ΔG and ΔG°?","acceptedAnswer":{"@type":"Answer","text":"ΔG describes free energy change under the actual conditions of the system, while ΔG° describes standard-state conditions. They connect through ΔG = ΔG° + RT ln Q."}},{"@type":"Question","@id":"https://fiveable.me/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr#how-is-ap-chemistry-95-tested","name":"How is AP Chemistry 9.5 tested?","acceptedAnswer":{"@type":"Answer","text":"AP Chemistry 9.5 is tested through sign and size reasoning with ΔG° and K, calculations using ΔG° = -RT ln K, unit conversions, and explanations of which side is favored at equilibrium."}}]}
```