---
title: "K = e^(-ΔG°/RT) — AP Chem Definition & Exam Guide"
description: "K = e^(-ΔG°/RT) links the equilibrium constant to standard free energy in AP Chem Topic 9.5. Learn why ΔG° < 0 means K > 1 and how the exam tests it."
canonical: "https://fiveable.me/ap-chem/key-terms/k-e-g-rt"
type: "key-term"
subject: "AP Chemistry"
unit: "Unit 9"
---

# K = e^(-ΔG°/RT) — AP Chem Definition & Exam Guide

## Definition

K = e^(-ΔG°/RT) is the AP Chem equation connecting the equilibrium constant (K) to standard free energy change (ΔG°), where R is the gas constant and T is temperature in Kelvin. If ΔG° < 0, then K > 1 and products are favored at equilibrium; if ΔG° > 0, then K < 1 and reactants are favored.

## What It Is

K = e^(-ΔG°/RT) is the bridge between thermodynamics and equilibrium. It takes ΔG°, the standard free energy change from [Unit 9](/ap-chem/unit-9 "fv-autolink"), and converts it into K, the equilibrium constant from Unit 7. It's the same relationship as [ΔG° = -RT ln K](/ap-chem/key-terms/g-rt-ln-k "fv-autolink"), just solved for K instead of ΔG°. In the equation, R is the gas constant (8.314 J/(mol·K)) and T is the absolute temperature in Kelvin.

The sign of ΔG° controls everything. A negative ΔG° makes the exponent positive, so K > 1 and products dominate at equilibrium. That's exactly what "[thermodynamically favored](/ap-chem/unit-9/thermodynamic-kinetic-control/study-guide/hRZ0V3goVueXCw1JeUdA "fv-autolink")" means in the CED (9.5.A.1). A positive ΔG° makes the exponent negative, so K < 1 and reactants dominate. And when ΔG° is near zero, K is close to 1, meaning neither side wins decisively. You rarely need to crunch the full exponential on the exam. What you do need is the qualitative logic: the bigger |ΔG°| is compared to RT, the more extremely K deviates from 1 (9.5.A.3).

## Why It Matters

This equation lives in Topic 9.5 (Free Energy and Equilibrium) in Unit 9: Thermodynamics and Electrochemistry, and it directly supports learning objective 9.5.A: explain whether a process is thermodynamically favored using the relationships between K, ΔG°, and T. It's listed verbatim in essential knowledge 9.5.A.2. Conceptually, it's the payoff of Unit 9. After spending the unit calculating ΔG° from enthalpy and entropy, this equation tells you what that number actually predicts, which is the composition of the [mixture](/ap-chem/key-terms/mixture "fv-autolink") once the reaction reaches equilibrium. It also closes a loop with [Unit 7](/ap-chem/unit-7 "fv-autolink"), because it explains where K values come from thermodynamically instead of just treating them as given numbers.

## Connections

### [ΔG° = -RT ln K (Unit 9)](/ap-chem/key-terms/g-rt-ln-k)

This is literally the same equation rearranged. Use ΔG° = -RT ln K when you know K and want free energy; use K = e^(-ΔG°/RT) when you know ΔG° and want K. The AP exam expects you to move fluently in both directions.

### [Spontaneity (Unit 9)](/ap-chem/key-terms/spontaneity)

This equation translates "thermodynamically favored" into equilibrium language. ΔG° < 0 doesn't mean the reaction is fast or goes to 100% completion. It means K > 1, so the equilibrium mixture contains more [products](/ap-chem/key-terms/products "fv-autolink") than reactants.

### [Law of Mass Action (Unit 7)](/ap-chem/key-terms/law-of-mass-action)

Unit 7 gives you K as a ratio of [concentrations](/ap-chem/unit-3/beer-lambert-law/study-guide/smCHzraorVz6qlWW1oeB "fv-autolink") from the mass action expression. Unit 9 tells you why K has the value it does. Together they let you compare Q to K and predict which direction a reaction shifts, then explain that shift thermodynamically.

### [Enthalpy (ΔH) (Units 6 & 9)](/ap-chem/key-terms/enthalpy-dh)

Since ΔG° = ΔH° - TΔS°, temperature changes ΔG°, which changes K through this equation. That's the thermodynamic reason behind Le Châtelier's temperature predictions, like why heating an [exothermic reaction](/ap-chem/key-terms/exothermic-reaction "fv-autolink") shrinks K.

## On the AP Exam

This shows up mostly as conceptual multiple-choice. A typical stem gives you a ΔG° value at 298 K, like ΔG° = -12 kJ/mol or +15.0 kJ/mol, and asks which statement about K or the reaction is true. You don't compute e^(-ΔG°/RT) by hand; you reason with signs and magnitudes. Negative ΔG° means K > 1, positive means K < 1, and ΔG° near zero means K near 1. Harder questions combine this with Q. For example, if ΔG° = +15.0 kJ/mol then K < 1, but if the mixture starts with Q = 0.01 and Q < K, the reaction still shifts toward products even though it's not thermodynamically favored overall. On FRQs, this relationship backs up justification prompts where you defend whether products or reactants are favored. Watch the units: ΔG° is usually given in kJ/mol but R is 8.314 J/(mol·K), so convert before plugging in.

## K = e^(-ΔG°/RT) vs Rate constant (k)

Capital K and lowercase k are completely different quantities. K is the equilibrium constant, set by thermodynamics through K = e^(-ΔG°/RT), and tells you how far a reaction goes. Lowercase k is the rate constant from kinetics (Unit 5) and tells you how fast it gets there. A reaction can have a huge K (very favored) and a tiny k (painfully slow). Diamond turning to graphite is favored thermodynamically but takes essentially forever.

## Key Takeaways

- K = e^(-ΔG°/RT) and ΔG° = -RT ln K are the same relationship rearranged, and you should be able to use either form.
- A negative ΔG° gives K > 1 (products favored at equilibrium), which is exactly what 'thermodynamically favored' means in the CED.
- A positive ΔG° gives K < 1 (reactants favored), and ΔG° near zero gives K near 1.
- The larger |ΔG°| is compared to RT, the more dramatically K deviates from 1 in either direction.
- ΔG° says nothing about speed; a thermodynamically favored reaction (K > 1) can still be extremely slow if its rate constant k is small.
- Always convert ΔG° from kJ/mol to J/mol before using R = 8.314 J/(mol·K), or the exponent will be off by a factor of 1000.

## FAQs

### What is K = e^(-ΔG°/RT) in AP Chem?

It's the equation in Topic 9.5 that relates the equilibrium constant K to the standard free energy change ΔG°, with R = 8.314 J/(mol·K) and T in Kelvin. It tells you that the sign and size of ΔG° determine whether K is greater than, less than, or close to 1.

### Does ΔG° < 0 mean the reaction goes to completion?

No. ΔG° < 0 means K > 1, so products are favored at equilibrium, but some reactants remain unless K is enormous. For example, ΔG° = -8.3 kJ/mol at 298 K gives a K only modestly above 1, not a reaction that runs to 100% completion.

### What's the difference between K and k in this equation?

Capital K is the equilibrium constant, determined by thermodynamics through ΔG°. Lowercase k is the rate constant from kinetics in Unit 5. K tells you how far a reaction proceeds; k tells you how fast. The two are independent of each other.

### Can a reaction with positive ΔG° still make products?

Yes. Positive ΔG° just means K < 1. If the current reaction quotient Q is even smaller than K (say Q = 0.01 when ΔG° = +15.0 kJ/mol at 298 K), the reaction still shifts toward products until Q rises to meet K.

### Do I have to calculate e^(-ΔG°/RT) on the AP exam?

Usually not. Most questions test qualitative reasoning, like recognizing that ΔG° = -12 kJ/mol means K > 1. If you do calculate, the equation is on the AP formula sheet, but remember to convert ΔG° to J/mol so its units match R.

## Related Study Guides

- [9.5 Free Energy and Equilibrium](/ap-chem/unit-9/free-energy-equilibrium/study-guide/ConxkA0PUZM7vwf8nPNr)

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