When a reversible reaction reaches equilibrium, the forward and reverse reactions run at the same rate and the concentrations stop changing. They do not have to be equal, but at a given temperature they always satisfy one fixed ratio: the equilibrium constant. For aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ), with each equilibrium concentration (mol/L) raised to the power of its coefficient in the balanced equation. Pure solids and liquids, and the solvent in dilute solution, are left out because their activity is 1.
The size of K tells you where the equilibrium lies. K much larger than 1 means mostly products at equilibrium; K much smaller than 1 means mostly reactants; values near 1 mean both are present in similar amounts. K depends only on temperature – adding a catalyst, changing concentrations or (for most reactions) changing pressure shifts the mixture but not K. K is also tied to thermodynamics through ΔG° = −RT·ln K.
Choose up to three reactants and three products, enter the coefficients and equilibrium concentrations, and the calculator returns K. Switch the mode to find a missing equilibrium concentration from a known K instead.
How to use the Equilibrium Constant Calculator
- Calculate
- Equilibrium constant K
- Number of reactants
- Number of products
- Reactant 1 – coefficient
- Reactant 1 – equilibrium concentration
- Reactant 2 – coefficient
- Reactant 2 – equilibrium concentration
- Reactant 3 – coefficient
- Reactant 3 – equilibrium concentration
- Product 1 – coefficient
- Product 1 – equilibrium concentration
- Product 2 – coefficient
- Product 2 – equilibrium concentration
- Product 3 – coefficient
- Product 3 – equilibrium concentration
- Temperature (for ΔG°)
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Each concentration is raised to its stoichiometric coefficient; the product of the product terms is divided by the product of the reactant terms. In the reverse modes the same expression is solved for the missing concentration, and ΔG° = −RT·ln K uses the temperature you enter.
Worked example
For N₂ + 3 H₂ ⇌ 2 NH₃ with illustrative equilibrium concentrations [N₂] = 0.25 M, [H₂] = 0.75 M and [NH₃] = 0.050 M: Kc = 0.050² ÷ (0.25 × 0.75³) = 0.0025 ÷ 0.10547 = 0.0237. log K = −1.625, and at 298.15 K ΔG° = −RT ln K ≈ +9.28 kJ/mol, so reactants are favoured.
Working backwards with K = 0.0237, [H₂] = 0.75 M and [NH₃] = 0.050 M, the missing nitrogen concentration is [N₂] = 0.050² ÷ (0.0237 × 0.75³) ≈ 0.25 M.
Manipulating K
| Change to the equation | New constant |
|---|---|
| Reverse the reaction | 1 / K |
| Multiply all coefficients by n | Kⁿ |
| Divide all coefficients by 2 | √K |
| Add two reactions | K₁ × K₂ |
Tips
- Check the equation is balanced before using coefficients.
- Find the limiting reagent before calculating a yield.
- Keep temperatures in kelvin for rate and equilibrium equations.
- Everything is calculated in your browser.
Common mistakes to avoid
- Using the excess reagent to calculate theoretical yield.
- Including pure solids and liquids in equilibrium expressions.
- Entering temperatures in °C into the Arrhenius equation without converting.
Inputs explained
| Field | What to enter |
|---|---|
| Calculate | Choose one: Equilibrium constant K, Unknown concentration of reactant 1, Unknown concentration of product 1 |
| Equilibrium constant K | A number |
| Number of reactants | Choose one: 1, 2, 3 |
| Number of products | Choose one: 1, 2, 3 |
| Reactant 1 – coefficient | A number |
| Reactant 1 – equilibrium concentration | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Reactant 2 – coefficient | A number |
| Reactant 2 – equilibrium concentration | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Reactant 3 – coefficient | A number |
| Reactant 3 – equilibrium concentration | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Product 1 – coefficient | A number |
| Product 1 – equilibrium concentration | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Product 2 – coefficient | A number |
| Product 2 – equilibrium concentration | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Product 3 – coefficient | A number |
| Product 3 – equilibrium concentration | Units: mol/L (M), mmol/L (mM), µmol/L (µM), nmol/L (nM), mol/m³ |
| Temperature (for ΔG°) | Units: K, °C, °F |
What each result means
- Equilibrium constant Kc = [K used]
- Equilibrium concentration of reactant 1 = ([Product 1 – equilibrium concentration]^[Product 1 – coefficient]×[Other product terms]÷([Equilibrium constant K]×[Other reactant terms]))^(1÷[Reactant 1 – coefficient])
- Equilibrium concentration of product 1 = ([Equilibrium constant K]×[Reactant 1 – equilibrium concentration]^[Reactant 1 – coefficient]×[Other reactant terms]÷[Other product terms])^(1÷[Product 1 – coefficient])
- log₁₀ K = log10([K used])
- K of the reverse reaction (1/K) = 1÷[K used]
- Standard Gibbs energy change ΔG° = −RT ln K = -8.314462618×[Temperature (for ΔG°)]×ln([K used])÷1000
Formula
aA + bB ⇌ cC + dD: Kc = [C]ᶜ·[D]ᵈ ÷ ([A]ᵃ·[B]ᵇ) · ΔG° = −R·T·ln KWorked example
Inputs — Calculate: Equilibrium constant K; Number of reactants: 2; Number of products: 1; Reactant 1 – coefficient: 1; Reactant 1 – equilibrium concentration: 0.25 mol/L (M); Reactant 2 – coefficient: 3; Reactant 2 – equilibrium concentration: 0.75 mol/L (M); Product 1 – coefficient: 2; Product 1 – equilibrium concentration: 0.05 mol/L (M); Temperature (for ΔG°): 298.15 K.
Results — Equilibrium constant Kc = 0.023704; log₁₀ K = -1.6252; K of the reverse reaction (1/K) = 42.1875; Standard Gibbs energy change ΔG° = −RT ln K = 9.2766 kJ/mol; Position of equilibrium = Reactants favoured (K < 1).
Key terms
- Equilibrium constant (Kc)
- The ratio of product to reactant concentrations, each raised to its coefficient, at equilibrium.
- Dynamic equilibrium
- The state in which forward and reverse reaction rates are equal, so concentrations stay constant.
- Activity
- The effective concentration of a species; 1 for pure solids and liquids.
- Law of mass action
- The rule that the equilibrium expression is built from the balanced equation's coefficients.
Frequently asked questions
How do you write an equilibrium constant expression?
Products over reactants, each equilibrium concentration raised to its coefficient: for 2 SO₂ + O₂ ⇌ 2 SO₃, Kc = [SO₃]² ÷ ([SO₂]²[O₂]).
What does a large K mean?
K ≫ 1 means the equilibrium mixture is mostly products; K ≪ 1 means mostly reactants. K = 1000 at equilibrium means the product term is a thousand times the reactant term.
Are solids and liquids included in K?
No. Pure solids, pure liquids and the solvent of a dilute solution have activity 1 and are left out of the expression.
Does a catalyst change K?
No. A catalyst speeds up the forward and reverse reactions equally, so equilibrium is reached sooner but K stays the same. Only temperature changes K.
What happens to K if I reverse or double the equation?
Reversing gives 1/K; multiplying all coefficients by 2 squares K; halving them takes the square root.
How is K related to ΔG°?
ΔG° = −RT·ln K. At 25 °C, K = 10 corresponds to ΔG° = −5.71 kJ/mol and K = 0.1 to +5.71 kJ/mol.
Does K have units?
Strictly, thermodynamic K is dimensionless because it uses activities. In practice Kc is often quoted with units of M raised to the power Δn.
What formula does the Equilibrium Constant Calculator use?
aA + bB ⇌ cC + dD: Kc = [C]ᶜ·[D]ᵈ ÷ ([A]ᵃ·[B]ᵇ) · ΔG° = −R·T·ln K.