acalculator

What is the equilibrium constant?

List the reactants and products with their coefficients and equilibrium concentrations or partial pressures. The calculator gives Kc or Kp with the working, and converts between Kc and Kp at a temperature you give.

Your numbers

Units
Work from
Amounts as
Reactants and products
Row 1
Row 2
Leave out pure solids and liquids; their activity is 1.
Optional with amounts: give it only when every row is a gas.
Equilibrium constant
Kc = 164.063

The equilibrium constant is Kc = 164.063.

Working
Kc = 0.042 ÷ 0.016² = 164.063
Kc
164.063
Δn
−1

Equilibrium constant: Kc = 164.063. The equilibrium constant is Kc = 164.063.

How to calculate

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.

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.

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.

  • 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.

Machine-readable copies: Markdown, JSON.

Worked examples

Each example is checked against the calculator on every build.

  1. Work from Amounts, Amounts as mol/L (Kc), Reactants and products Reactant 2 0.016; Product 1 0.042 gives Kc 164.0625, Working Kc = 0.042 ÷ 0.016² = 164.063, Δn -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. Work from Amounts, Amounts as mol/L (Kc), Reactants and products Reactant 2 0.9; Reactant 1 0.35; Product 2 1.1 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. Work from Amounts, Amounts as mol/L (Kc), Reactants and products Reactant 1 0.011; Reactant 1 0.0011; Product 1 0.037; Product 1 0.046 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. Work from Kc ⇄ Kp, Amounts as mol/L (Kc), Known K 0.28, Δn (gas moles) -2, Temperature 1,173 K 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. Work from Kc ⇄ Kp, Amounts as mol/L (Kc), Known K 0.0952, Δn (gas moles) 2, Temperature 500 K 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

How it works

Equilibrium constant (OpenStax §13.2). For a balanced reaction aA + bB ⇌ cC + dD,

K = [C]^c × [D]^d ÷ ([A]^a × [B]^b)

Each row is one gas or dissolved species: reactant or product, its whole-number coefficient (1 to 20) and its equilibrium amount (above 0, up to 10⁶). With amounts in mol/L the result is Kc; in atm it is Kp. With no reactant rows the denominator is 1; with no product rows the numerator is 1. Pure solids and liquids are left out. Typed amounts are exact decimals, so K is exact until it is shown to 6 significant figures, trailing zeros dropped; a value below 0.000001 or of 10²¹ or more is written with a power of ten (Kc = 1.2 × 10⁻⁷ for one product at 0.00000012 mol/L), and a value from 0.000001 up to 10²¹ is written out (0.0000302223). The working shows each typed amount to 10 significant figures (with a power of ten below 0.000001, as for K) and its coefficient as a superscript (0.042 ÷ 0.016² = 164.063). An amount written with a power of ten is always put in brackets, so a coefficient is a power of all of it (1 ÷ (1 × 10⁻¹⁵)²⁰ = 1 × 10³⁰⁰).

Δn. The sum of the product coefficients minus the sum of the reactant coefficients, treating every row as a gas.

Kc and Kp.

Kp = Kc × (RT)^Δn, so Kc = Kp ÷ (RT)^Δn

with T in kelvins (°C + 273.15; °F converts exactly) and R = 8.31446261815324 ÷ 101.325 = 0.0820573660809597 L·atm/(mol·K), from the exact R in J/(mol·K) and 1 atm = 101.325 kPa. OpenStax rounds R to 0.0821, which moves Kp in the third significant figure.

  • Amounts: give a temperature only when every listed species is a gas, and the page also shows the other constant.
  • Kc ⇄ Kp: type the known constant (10⁻¹⁰⁰ to 10¹⁰⁰), Δn (a whole number from −40 to 40) and the temperature; the headline is the other constant.

A result too large or too small to hold as a number gives no answer.

Worked examples by hand

2NO₂ ⇌ N₂O₄ (OpenStax Example 13.2). Kc = 0.042 ÷ 0.016² = 0.042 ÷ 0.000256 = 164.0625 (OpenStax: 1.6 × 10²). Δn = 1 − 2 = −1.

2SO₂ + O₂ ⇌ 2SO₃ (Check Your Learning). Kc = 1.1² ÷ (0.90² × 0.35) = 1.21 ÷ 0.2835 = 4.26808.

CO + H₂O ⇌ CO₂ + H₂, experiment 2 (Example 13.3). Q = (0.037 × 0.046) ÷ (0.011 × 0.0011) = 0.001702 ÷ 0.0000121 = 140.661, above K = 0.64, so the reaction runs in reverse.

Kc = 0.28 at 1173 K, Δn = −2 (Example 13.4 d). RT = 0.0820573660809597 × 1173 = 96.2533; Kp = 0.28 ÷ 96.2533² = 0.0000302223, which is 3.02223 × 10⁻⁵ (OpenStax: 3.0 × 10⁻⁵).

Kc = 0.0952 at 500 K, Δn = 2. RT = 41.0287; Kp = 0.0952 × 41.0287² = 160.255 (OpenStax: 160).

Other questions people ask

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.