ΔG° = -RT ln K in AP Chemistry

ΔG° = -RT ln K is the AP Chem equation connecting standard free energy change to the equilibrium constant, where R is the gas constant (8.314 J/mol·K) and T is temperature in Kelvin. A negative ΔG° means K > 1 and products are favored; a positive ΔG° means K < 1 and reactants are favored.

What is ΔG° = -RT ln K?

ΔG° = -RT ln K is the bridge between thermodynamics and equilibrium. On one side you have ΔG°, the standard free energy change, which tells you whether a reaction is thermodynamically favored. On the other side you have K, the equilibrium constant, which tells you where the reaction actually ends up. This equation says those are the same piece of information written two different ways.

The negative sign and the natural log do all the interesting work. When K > 1 (products favored), ln K is positive, so ΔG° comes out negative, which is the AP definition of "thermodynamically favored" (EK 9.5.A.1). When K < 1, ln K is negative and ΔG° is positive, so reactants win. When K is close to 1, ΔG° is close to zero. R is the gas constant, 8.314 J/(mol·K) on your equation sheet, and T must be in Kelvin. The rearranged form K = e^(-ΔG°/RT) lets you go the other direction, solving for K when you're given ΔG°.

Why ΔG° = -RT ln K matters in AP® Chemistry

This equation lives in Topic 9.5 (Free Energy and Equilibrium) in Unit 9: Thermodynamics and Electrochemistry. It directly supports learning objective 9.5.A, which asks you to explain whether a process is thermodynamically favored using the relationships between K, ΔG°, and T. The CED explicitly lists both ΔG° = -RT ln K and K = e^(-ΔG°/RT) as required equations (EK 9.5.A.2), and it also expects qualitative estimation (EK 9.5.A.3). That means you need both skills, plugging in numbers correctly AND eyeballing a K value to predict the sign of ΔG° without a calculator. This equation is also the conceptual payoff of Unit 9. It ties everything you learned about equilibrium in Units 7 and 8 back to energy, answering the question "WHY does K have the value it has?"

How ΔG° = -RT ln K connects across the course

K = e^(-ΔG°/RT) (Unit 9)

This is the same equation solved for K instead of ΔG°. You use this form when a problem hands you a free energy value and asks where equilibrium lies. Knowing both directions means one equation covers twice the question types.

Spontaneity (Unit 9)

ΔG° < 0 is the math behind the phrase "thermodynamically favored." This equation translates that energy statement into an equilibrium statement, since a negative ΔG° forces K > 1. Favored isn't a vibe, it's a number.

Law of Mass Action (Unit 7)

The law of mass action defines K from concentrations or pressures back in Unit 7. ΔG° = -RT ln K explains where that K comes from energetically. Unit 7 measures equilibrium; Unit 9 explains it.

Enthalpy (ΔH) (Units 6 & 9)

You usually get ΔG° from ΔG° = ΔH° - TΔS° first, then feed it into -RT ln K to find K. Chained together, these two equations let you predict an equilibrium constant from a reaction's heat and entropy data.

Rate constant (k) (Unit 5)

Lowercase k is kinetics (how fast), capital K is equilibrium (how far). ΔG° = -RT ln K says nothing about speed. A reaction with a huge K and very negative ΔG° can still be unmeasurably slow if the activation energy is high.

Is ΔG° = -RT ln K on the AP® Chemistry exam?

This shows up as a calculation and as a sign-logic check. Calculation version: you're given K at a temperature (usually 298 K) and asked for ΔG°, like K = 3.5 × 10⁻⁵ or K = 3.5 × 10⁵ at 298 K. The classic trap is units, since R = 8.314 J/(mol·K) gives you an answer in J/mol, but answer choices are usually in kJ/mol. Divide by 1000 or you'll pick the wrong choice. Sign-logic version: given K = 0.025, which statements must be true? You should immediately think K < 1, so ln K < 0, so ΔG° > 0, so reactants are favored. Or given ΔG° = -15.7 kJ/mol, conclude K > 1 and products dominate at equilibrium. On FRQs, this equation often appears as one step in a longer thermodynamics problem, frequently after you've calculated ΔG° from ΔH° and ΔS°. Both forms of the equation are on the AP equation sheet, so memorize the logic, not the formula.

ΔG° = -RT ln K vs ΔG (non-standard free energy change)

ΔG° (with the degree symbol) is the free energy change under standard conditions and is a fixed value for a reaction at a given temperature. It pairs with K in this equation. Plain ΔG depends on the actual current conditions and pairs with Q instead. At equilibrium, ΔG = 0, but ΔG° is almost never zero. If you write "ΔG° = 0 at equilibrium" on an FRQ, that's a lost point. ΔG° = 0 only in the special case where K = 1.

Key things to remember about ΔG° = -RT ln K

• ΔG° = -RT ln K connects standard free energy change to the equilibrium constant, so knowing one tells you the other.

• A negative ΔG° means K > 1 and products are favored at equilibrium; a positive ΔG° means K < 1 and reactants are favored.

• When ΔG° is near zero, K is near 1; when ΔG° is much bigger than RT in magnitude, K is far from 1.

• R = 8.314 J/(mol·K) gives ΔG° in joules per mole, so convert to kJ/mol before matching answer choices.

• T must always be in Kelvin, so convert 25°C to 298 K before plugging in.

• ΔG° = 0 does not happen at equilibrium in general; ΔG = 0 at equilibrium, while ΔG° = 0 only when K = 1.

Frequently asked questions about ΔG° = -RT ln K

What is ΔG° = -RT ln K and what does each variable mean?

It's the equation linking standard free energy change (ΔG°) to the equilibrium constant (K), where R is the gas constant (8.314 J/mol·K) and T is absolute temperature in Kelvin. It's required for Topic 9.5 in AP Chem and appears on the equation sheet.

Does ΔG° = 0 at equilibrium?

No, and this is a classic exam trap. ΔG (no degree symbol) equals zero at equilibrium. ΔG° is a fixed property of the reaction and only equals zero in the special case where K = 1.

What's the difference between K in ΔG° = -RT ln K and the rate constant k?

Capital K is the equilibrium constant from Unit 7 and tells you how far a reaction goes. Lowercase k is the rate constant from Unit 5 kinetics and tells you how fast it goes. ΔG° relates only to capital K; a thermodynamically favored reaction can still be extremely slow.

If K is less than 1, is ΔG° positive or negative?

Positive. When K < 1, ln K is negative, and the negative sign in front of RT flips it positive. For example, K = 0.025 at 25°C gives ΔG° ≈ +9.1 kJ/mol, meaning reactants are favored.

Why does my ΔG° answer come out 1000 times too big?

Because R = 8.314 J/(mol·K) gives an answer in J/mol, while AP answer choices are usually in kJ/mol. Divide by 1000. For K = 3.5 × 10⁻⁵ at 298 K, you get about +25,400 J/mol, which is +25.4 kJ/mol.