# What is the equilibrium constant?

Finds the equilibrium constant Kc or Kp of a reaction from the equilibrium concentrations or partial pressures and the balanced coefficients, and converts between Kc and Kp with Kp = Kc(RT)^Δn.

- Page: https://www.acalculator.org/chemistry/equilibrium-constant-calculator
- JSON spec: https://www.acalculator.org/chemistry/equilibrium-constant-calculator.json
- Version: ce4fb2aa6fa4

## Default answer

Example with the default inputs (Work from Amounts, Amounts as mol/L (Kc), Reactants and products [Side Reactant, Coefficient 2, Amount 0.016; Side Product, Coefficient 1, Amount 0.042]): The equilibrium constant is Kc = 164.063.

## Inputs

| Key | Label | Description |
| --- | --- | --- |
| m | Work from | Equilibrium amounts of every species, or a known Kc or Kp to convert. |
| k | Amounts as | Concentrations in mol/L give Kc; partial pressures in atm give Kp. In Kc ⇄ Kp, the constant you know. |
| species | Reactants and products | One row per gas or dissolved species: reactant or product, its coefficient, and its equilibrium amount. |
| kv | Known K | The equilibrium constant you know: Kc or Kp, as chosen above. |
| dn | Δn (gas moles) | Moles of gaseous products minus moles of gaseous reactants in the balanced equation. |
| t | Temperature | The temperature, to convert between Kc and Kp; only when every listed species is a gas. |

## Outputs

| Key | Label | Description |
| --- | --- | --- |
| result | Equilibrium constant | Kc or Kp, to 6 significant figures. |
| expression | Working | Products’ amounts to their coefficients over reactants’ amounts to theirs. |
| kc | Kc | The equilibrium constant in concentrations (mol/L). |
| kp | Kp | The equilibrium constant in partial pressures (atm). |
| dnOut | Δn | Moles of gaseous products minus moles of gaseous reactants. |
| other | Converted | The other constant from Kp = Kc × (RT)^Δn, with R = 0.0820574 L·atm/(mol·K). |

## Method

K = product of (product amounts ^ coefficients) ÷ product of (reactant amounts ^ coefficients); Kp = Kc × (RT)^Δn with R = 8.31446261815324 ÷ 101.325 = 0.0820573660809597 L·atm/(mol·K) and Δn = gas moles of products − gas moles of reactants.

## Assumptions

- The amounts are equilibrium values; amounts away from equilibrium give the reaction quotient Q instead.
- Pure solids and liquids are left out (their activity is 1); concentrations and pressures stand for activities.
- Converting between Kc and Kp treats every listed species as a gas; Δn counts gas moles only.

## Worked examples

1. m = amounts, k = c, species = {"side":"r","n":2,"x":0.016} or {"side":"p","n":1,"x":0.042} gives kc = 164.0625, expression = Kc = 0.042 ÷ 0.016² = 164.063, dnOut = -1. Source: OpenStax, Chemistry 2e, §13.2 Equilibrium Constants (Qc and Kc; Kp = Kc(RT)^Δn), https://openstax.org/books/chemistry-2e/pages/13-2-equilibrium-constants, Example 13.2 (b): Kc = 0.042 ÷ (0.016)² = 1.6 × 10².
2. m = amounts, k = c, species = {"side":"r","n":2,"x":0.9} or {"side":"r","n":1,"x":0.35} gives kc = 4.268078. Source: OpenStax, Chemistry 2e, §13.2 Equilibrium Constants (Qc and Kc; Kp = Kc(RT)^Δn), https://openstax.org/books/chemistry-2e/pages/13-2-equilibrium-constants, Check Your Learning after Example 13.2: Kc = 4.3.
3. m = amounts, k = c, species = {"side":"r","n":1,"x":0.011} or {"side":"r","n":1,"x":0.0011} gives kc = 140.661157. Source: OpenStax, Chemistry 2e, §13.2 Equilibrium Constants (Qc and Kc; Kp = Kc(RT)^Δn), https://openstax.org/books/chemistry-2e/pages/13-2-equilibrium-constants, Example 13.3 Predicting the Direction of Reaction, experiment 2: Qc = 1.4 × 10².
4. m = convert, k = c, kv = 0.28, dn = -2, t = 1,173 gives kp = 0.00003. Source: OpenStax, Chemistry 2e, §13.2 Equilibrium Constants (Qc and Kc; Kp = Kc(RT)^Δn), https://openstax.org/books/chemistry-2e/pages/13-2-equilibrium-constants, Example 13.4 (d): Kp = 0.28 × [(0.0821)(1173)]⁻² = 3.0 × 10⁻⁵.
5. m = convert, k = c, kv = 0.0952, dn = 2, t = 500 gives kp = 160.25519. Source: OpenStax, Chemistry 2e, §13.2 Equilibrium Constants (Qc and Kc; Kp = Kc(RT)^Δn), https://openstax.org/books/chemistry-2e/pages/13-2-equilibrium-constants, Check Your Learning after Example 13.4 (d): Kp = 0.0952 × [(0.0821)(500)]² = 160.

## FAQ

### How do I calculate the equilibrium constant?

For aA + bB ⇌ cC + dD, K = [C]^c × [D]^d ÷ ([A]^a × [B]^b): raise each equilibrium amount to its coefficient, multiply the products’ terms, and divide by the reactants’ terms. For 2NO₂ ⇌ N₂O₄ with [NO₂] = 0.016 M and [N₂O₄] = 0.042 M, Kc = 0.042 ÷ 0.016² = 164.

### What is the difference between Kc and Kp?

Kc uses molar concentrations (mol/L); Kp uses partial pressures of gases (atm). They are related by Kp = Kc × (RT)^Δn, where Δn is the moles of gaseous products minus the moles of gaseous reactants and R = 0.08206 L·atm/(mol·K). When Δn = 0, Kp = Kc.

### Why are solids and liquids left out?

The relative concentration of a pure solid or liquid is 1, so it does not change the product; only gases and dissolved species appear in K. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp is just the pressure of CO₂.

### What does the size of K tell me?

K above 1 means products are more abundant at equilibrium, K below 1 means reactants are, and K near 1 means comparable amounts. K depends on temperature, so quote it with the temperature.

### What if my amounts are not at equilibrium?

Then the same formula gives the reaction quotient Q. Compare it with K: if Q < K the reaction runs forward, if Q > K it runs in reverse. For CO + H₂O ⇌ CO₂ + H₂ with K = 0.64, the experiment-2 amounts give Q = 141, so the reaction shifts back toward CO and H₂O.

### How do I find Δn?

Add the coefficients of the gaseous products and subtract those of the gaseous reactants. For N₂ + 3H₂ ⇌ 2NH₃, Δn = 2 − 4 = −2, so Kp = Kc ÷ (RT)². With amounts, the page works Δn out from the rows, treating every row as a gas.

## Sources

- OpenStax, Chemistry 2e, §13.2 Equilibrium Constants: Qc and Kc for mA + nB ⇌ xC + yD, Qp, Kp = Kc(RT)^Δn, solids and liquids left out, Example 13.2 (2NO₂ ⇌ N₂O₄: Kc = 1.6 × 10²) and its Check Your Learning (Kc = 4.3), Example 13.3 (experiment 2: Qc = 1.4 × 10², reverse), Example 13.4 (d) (Kp = 3.0 × 10⁻⁵) and its Check Your Learning (d) (Kp = 160). https://openstax.org/books/chemistry-2e/pages/13-2-equilibrium-constants (retrieved 2026-10-19)
- OpenStax, Chemistry 2e, §16.4 Free Energy, Table 16.4: K above 1, products more abundant; below 1, reactants; equal to 1, comparable. https://openstax.org/books/chemistry-2e/pages/16-4-free-energy (retrieved 2026-10-19)
- NIST, CODATA 2022 value: molar gas constant R = 8.314 462 618… J/(mol·K), exact. https://physics.nist.gov/cgi-bin/cuu/Value?r (retrieved 2026-10-19)