Equilibrium Constant (Keq) from Gibbs Free Energy

Free Equilibrium Constant (K) from Gibbs Free Energy Calculator

This calculator finds the equilibrium constant, K, directly from a reaction's standard Gibbs free energy change, ΔG°, using the equation ΔG° = −RT ln K. Enter ΔG° and a temperature and it instantly returns K, along with a plain-language read on whether the reaction is spontaneous and how far it runs toward products at equilibrium.

It also works in reverse. If you already know K and want the standard free energy behind it, or if you know both ΔG° and K and need to find the temperature at which that relationship holds, just switch the 'solve for' dropdown. A second Advanced Tools mode converts between Kp and Kc for gas-phase reactions, using the ideal-gas relationship Kp = Kc(RT)^Δn — a step that trips up a lot of general chemistry students. Every result comes with a full worked solution, an equilibrium-scale diagram, and instant answers, completely free with no sign-up.

What Links Gibbs Free Energy to the Equilibrium Constant?

Every reaction has a standard Gibbs free energy change, ΔG°, that tells you how favorable it is under standard reference conditions — 1 M concentrations for solutions, 1 atm for gases, and a chosen temperature (usually 298.15 K). Separately, every reaction also has an equilibrium constant, K, that describes the actual ratio of products to reactants once the reaction has settled down and stopped changing.

These two numbers aren't independent — they're two different descriptions of the exact same chemical fact, connected by ΔG° = −RT ln K, where R is the gas constant, 8.314 J/(mol·K), and T is the absolute temperature in kelvin. Knowing one always lets you calculate the other, which is exactly what this calculator automates.

How to Read the Result: What Does K Actually Mean?

A large K (say, above 1,000) means the reaction proceeds almost entirely to products — at equilibrium, there's very little of the starting material left. A small K (below 0.001) means the opposite: the reaction barely gets going, and the mixture at equilibrium is overwhelmingly reactants. A K close to 1 means neither side wins outright — a genuinely mixed equilibrium with meaningful amounts of both reactants and products present.

The sign of ΔG° tells the same story from a different angle. A negative ΔG° (spontaneous under standard conditions) always pairs with K greater than 1. A positive ΔG° (non-spontaneous under standard conditions) always pairs with K less than 1. A ΔG° of exactly zero means K equals exactly 1 — the reaction sits precisely balanced between reactants and products.

How to Use This Calculator

On the Standard Solver tab, pick what you're solving for. Solving for K (the default and most common use) needs ΔG° — pick your preferred unit (J/mol, kJ/mol, cal/mol, or kcal/mol) — and a temperature, which can be entered in kelvin, °C, or °F. Solving for ΔG° instead needs a known K and temperature. Solving for temperature needs both ΔG° and K, and is useful when a lab result gives you both values and asks at what temperature they'd actually be consistent.

The Advanced Tools tab handles gas-phase equilibria specifically, where chemists often need to switch between Kp (using partial pressures in atmospheres) and Kc (using molar concentrations). Enter either constant, a temperature, and Δn — the change in moles of gas between products and reactants in the balanced equation — and the calculator converts instantly using Kp = Kc(RT)^Δn.

Worked Example: Finding K from ΔG°

A reaction has a standard free energy change of ΔG° = −28.6 kJ/mol at 298.15 K. What is its equilibrium constant?

First convert to base units: ΔG° = −28,600 J/mol. Then apply K = e^(−ΔG°/RT) = e^(−(−28,600) / (8.314 × 298.15)) = e^(28,600 / 2,478.8) = e^11.54 ≈ 102,600.

A K this large means the reaction runs almost entirely to products — at equilibrium, essentially no meaningful amount of starting material remains, which matches the strongly negative (spontaneous) ΔG° used to calculate it.

Worked Example: Finding ΔG° from a Measured K

A lab measurement finds an equilibrium constant of K = 4.7 × 10⁻³ for a reaction at 350 K. What is the standard free energy change?

Apply ΔG° = −RT ln K = −8.314 × 350 × ln(0.0047) = −8.314 × 350 × (−5.36) = +15,600 J/mol, or about +15.6 kJ/mol.

The positive ΔG° confirms what the small K already suggested: this reaction is non-spontaneous under standard conditions and favors the reactants at equilibrium.

Worked Example: Converting Kc to Kp for a Gas Reaction

For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K, Kc = 0.061. What is Kp? Here, Δn = (moles of gas products) − (moles of gas reactants) = 2 − (1 + 3) = −2.

Apply Kp = Kc(RT)^Δn with R = 0.08206 L·atm/(mol·K): RT = 0.08206 × 500 = 41.03. So Kp = 0.061 × (41.03)⁻² = 0.061 / 1,683.5 ≈ 3.62 × 10⁻⁵.

Notice Kp came out much smaller than Kc here — that's expected whenever Δn is negative, since (RT)^Δn shrinks the value rather than growing it. Getting the sign of Δn right is the single most common mistake in this conversion.

Standard Conditions vs Real Conditions: Why ΔG° Isn't the Whole Story

ΔG° and the K calculated from it both describe standard reference conditions — a snapshot, not the whole reaction as it actually runs in a flask. At any real moment, a reaction's true driving force is ΔG = ΔG° + RT ln Q, where Q is the reaction quotient, calculated the same way as K but using whatever concentrations exist right now rather than the ones at equilibrium.

This calculator focuses specifically on the ΔG°–K relationship at equilibrium. For adjusting to non-standard concentrations with the reaction quotient Q, or for converting a free energy value into an electrochemical cell potential E°, see the Gibbs Free Energy Calculator's Advanced Tools tab.

Does K Change with Temperature?

Yes — K is only constant at a fixed temperature. Change the temperature and K changes too, because both ΔH° and ΔS° (which together make up ΔG°) contribute differently at different temperatures. For an exothermic reaction, K generally falls as temperature rises; for an endothermic reaction, K generally rises. The van 't Hoff equation describes exactly how K shifts between two temperatures and is the right tool once temperature dependence itself is the question, rather than a single ΔG°–K–T conversion at one temperature, which is what this calculator handles.

Kp vs Kc: When Each One Is Used

Kc is defined using equilibrium molar concentrations, [products]/[reactants], each raised to its stoichiometric coefficient — the natural choice for reactions in solution. Kp uses partial pressures in atmospheres instead, and is the natural choice for pure gas-phase reactions, since pressure is usually what's actually measured in a gas-phase experiment.

Whenever the total number of gas moles changes during a reaction (Δn ≠ 0), Kp and Kc take different numerical values, related by Kp = Kc(RT)^Δn. When Δn = 0 — the same number of gas moles on both sides — Kp and Kc are numerically identical, since (RT)⁰ = 1.

Common Mistakes to Avoid

A few small errors account for most wrong answers when moving between ΔG°, K, and temperature.

• Forgetting to convert temperature to kelvin — R = 8.314 J/(mol·K) only works with absolute temperature, never °C or °F directly.
• Mixing energy units mid-calculation — keep ΔG° in J/mol (not kJ/mol) when substituting directly into ΔG° = −RT ln K, or convert consistently throughout.
• Getting the sign of Δn backwards in the Kp/Kc conversion — Δn is moles of gas products minus moles of gas reactants, and getting this flipped inverts the whole conversion factor.
• Trying to take ln(K) when K is zero or negative — K is always a positive number, since it comes from concentrations or pressures, which can't be negative.
• Assuming K calculated from ΔG° applies at a different temperature — K is only valid at the exact temperature used to calculate it.

Where This Calculation Shows Up in Real Life

Finding K from ΔG° (and back again) underpins predicting whether an industrial reaction is worth running at a given temperature, estimating drug-receptor binding strength from measured free energies in pharmacology, calculating gas-phase equilibrium yields for reactions like ammonia synthesis, and checking whether a proposed reaction pathway in a synthesis route is thermodynamically favorable before ever running it in a lab.

Limitations to Keep in Mind

This calculator assumes ideal behavior — ideal solutions for Kc and ideal gases for Kp — and a single fixed temperature. Real systems at high concentration or high pressure deviate from ideal behavior, and K calculated this way is always the thermodynamic (standard-state) value, not necessarily what a specific real mixture will show if the reaction hasn't actually reached equilibrium yet. For precision work, always compare against measured, published equilibrium data where available.

Quick Reference: Every Formula on This Page

ΔG° = −RT ln K, and equivalently K = e^(−ΔG°/RT) — the core link between standard free energy and the equilibrium constant. R = 8.314 J/(mol·K) — the gas constant used here. Kp = Kc × (RT)^Δn, using R = 0.08206 L·atm/(mol·K) — the conversion between gas-phase equilibrium constants. Δn = moles of gas products − moles of gas reactants, from the balanced chemical equation.