Equilibrium Constant Calculator

Enter coefficients and equilibrium concentrations for each reactant and product.

Science log K ΔG° up to 4+4 species
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Equilibrium constant calculator

K = [products] / [reactants]

Instructions — Equilibrium Constant Calculator

Set up your reactant side and product side independently. For each species enter the stoichiometric coefficient, a short name (for display), and the equilibrium concentration in M (mol/L). Add or remove rows as needed — the tool supports up to 4 reactants and 4 products.

The result panel shows:

  • The equilibrium constant K and the symbolic expression
  • log10 K and ln K
  • The standard Gibbs free energy ΔG° at the temperature you choose (default 298.15 K), computed from ΔG° = −RT ln K
  • Δn — the difference between total product and reactant moles, useful for converting Kc to Kp

Formulas

For a generic reaction aA + bB ↔ cC + dD at equilibrium:

Kc = ([C]c · [D]d) / ([A]a · [B]b)

Pressure form (gases): Kp = Kc (RT)Δn   where Δn = (c+d) − (a+b)

Link to Gibbs energy: ΔG° = −RT ln K

Reaction quotient: Q has the same form as K but uses non-equilibrium concentrations. Q < K shifts forward; Q > K shifts back; Q = K means already at equilibrium.

Reference

  • Kw (water autoionization, 25 °C) = 1.0 × 10−14
  • Ka acetic acid = 1.75 × 10−5 — weak acid, large pKa
  • Ka HCl ≈ 106 — strong acid, essentially full dissociation
  • N2 + 3 H2 ↔ 2 NH3 at 298 K, K ≈ 6 × 105 — thermodynamically favored, but kinetically slow at low T
  • 2 NO2 ↔ N2O4 at 298 K, K ≈ 8.8 atm−1 — classic Le Chatelier demonstration

K only changes with temperature (van ’t Hoff equation). Concentration, pressure, and catalysts shift the position of equilibrium but leave K untouched.

Article — Equilibrium Constant Calculator

How the equilibrium constant calculator works

The equilibrium constant K describes the ratio of product to reactant activities (or concentrations, pressures) at equilibrium for a balanced chemical reaction. A large K means products are favored; a small K means reactants dominate. This calculator handles up to four reactants and four products, returns K, log K, ln K, and the Gibbs free energy ΔG° = −RT ln K at the temperature you set.

K is the bridge between thermodynamics and analytical chemistry. From a single number you can predict reaction direction, equilibrium composition, pH of weak acid solutions, and solubility of sparingly soluble salts.

What is the equilibrium constant

Chemical equilibrium is the state where the forward and reverse rates of a reversible reaction are equal. Concentrations stop changing, but the reaction has not stopped — it is dynamic. K is the ratio that the system arrives at, fixed by temperature alone.

For aA + bB ↔ cC + dD with all species in solution:

K = [C]c[D]d / ([A]a[B]b)

Pure solids and pure liquids do not appear in the expression because their activities are 1. This is why dissolution and decomposition equilibria like CaCO3(s) ↔ CaO(s) + CO2(g) reduce to K = P(CO2).

Types of equilibrium constants

Different reactions need different K subscripts. They all share the same mathematical form but differ in which quantities go in.

K types
Kc molar concentrations
Kp partial pressures (gas)
Ksp solubility product
Ka, Kb weak acid / base dissociation
Kw water autoionization (10−14)
Kf complex ion formation

Convert Kc to Kp using Kp = Kc(RT)Δn, where Δn is the change in moles of gas (products minus reactants). For reactions with no net gas change, Kc and Kp are numerically equal.

The equilibrium constant formula

The calculator builds the expression from your stoichiometric coefficients automatically. For the Haber process N2 + 3 H2 ↔ 2 NH3:

K = [NH3]2 / ([N2] · [H2]3)

With [N2] = 0.1 M, [H2] = 0.2 M, [NH3] = 0.4 M, the numerator is 0.16 and the denominator is 0.1 × 0.008 = 0.0008. K = 200. Log K = 2.3 and ΔG° at 298 K is about −13 kJ/mol — comfortably product-favored at the operating temperature.

Did you know

The Haber-Bosch process for industrial ammonia synthesis runs at 450 °C and 150–300 atm. The high pressure shifts equilibrium toward NH3 (Le Chatelier); the high temperature is needed to make the reaction fast despite the unfavorable shift at high T. The compromise is one of the great engineering achievements of the 20th century.

Temperature and the equilibrium constant

K changes only with temperature. Concentration, pressure, and catalysts all shift the position of equilibrium but never change K itself. The van 't Hoff equation describes the temperature dependence:

ln(K2/K1) = −(ΔH°/R)(1/T2 − 1/T1)

For an exothermic reaction (ΔH° < 0), K decreases with rising T. For an endothermic reaction K increases. The relation gives you K at any T2 once you know K at T1 and the enthalpy of reaction.

The ICE table method

ICE stands for Initial, Change, and Equilibrium. It is the standard technique for solving equilibrium problems when you start with non-equilibrium concentrations and want to find the equilibrium state.

  • Initial row — the starting concentrations of every species
  • Change row — the algebraic change, with x as the unknown extent of reaction. Reactants change by −coef · x, products by +coef · x.
  • Equilibrium row — initial + change. These go into the K expression.

You solve the resulting polynomial for x (often quadratic, occasionally cubic), then plug back to get all equilibrium concentrations. Once you have those concentrations, this calculator returns K instantly.

Equilibrium constant vs reaction quotient

The reaction quotient Q is calculated with the same formula as K but using current (not necessarily equilibrium) concentrations. Comparing Q to K tells you which way the reaction will shift:

Q versus K
Q < K reaction proceeds forward
Q > K reaction proceeds backward
Q = K already at equilibrium

This is operationally how Le Chatelier's principle is justified. Add more reactant, Q drops below K, reaction shifts forward until Q catches up. Remove product, same outcome. Change temperature, K itself moves, and the system relaxes to the new K.

Weak acid equilibrium constants

Weak acids partially dissociate in water. The dissociation constant Ka measures how partial: acetic acid has Ka = 1.75 × 10−5 (pKa = 4.76), so a 0.1 M solution dissociates only about 1.3% into ions. Stronger weak acids like HF (Ka = 6.6 × 10−4, pKa = 3.18) dissociate 7% at the same concentration; weaker ones like HCN (Ka = 6.2 × 10−10, pKa = 9.21) barely ionize at all.

The pH of a weak acid solution drops out of the equilibrium expression. For monoprotic HA with initial concentration c, applying the small-x approximation gives [H+] ≈ √(Ka × c). A 0.1 M acetic acid solution has pH around 2.87. The same logic applies to weak bases through Kb, with the pH lying above 7.

Common equilibrium constant mistakes

K is dimensionless — but the underlying units matter

Strictly K is a ratio of activities (dimensionless). In practice we use concentrations or partial pressures and the K value carries implicit units that depend on the stoichiometry. Comparing K values across reactions with different Δn requires care — convert to a common reference (often Kp at 1 atm or Kc at 1 M).

  • Including pure solids or liquids — their activity is 1 and they drop out of K
  • Wrong sign of Δn when converting Kc to Kp — Δn = moles gas products − moles gas reactants
  • Forgetting that K only changes with T — concentration changes shift position, not K
  • Using initial concentrations instead of equilibrium concentrations in the K expression
  • Mixing up Ka and pKa — pKa = −log Ka; small Ka means large pKa
Tip

For weak acids with Ka below about 10−4, the x produced by dissociation is small compared with the initial acid concentration. Approximate (HA − x) ≈ HA and skip the quadratic. The error is usually under 5% and saves the algebra.

For solubility-product problems involving sparingly soluble salts, the Ksp form follows the same pattern. AgCl dissolves to give Ag+ and Cl; Ksp = [Ag+][Cl] = 1.8 × 10−10. From that you can compute molar solubility in pure water as √Ksp = 1.34 × 10−5 M. Adding a common ion (NaCl) reduces solubility by Le Chatelier; adding a complexing agent (NH3) can shift K and increase solubility dramatically.

FAQ

Their activity is taken as 1 and does not appear in the equilibrium expression. The concentration of a pure solid or pure liquid is fixed by its density and does not change as the reaction proceeds, so it cannot affect K. That is why CaCO3(s) and the H2O(l) in some weak-acid problems disappear from the formula.
K_c uses molar concentrations and K_p uses partial pressures (atmospheres or bar). They are linked by K_p = K_c (RT)^Delta-n, where Delta-n counts gas-phase moles only. For a reaction with no change in gas moles, K_c and K_p are numerically equal.
The van’t Hoff equation gives ln(K2/K1) = -Delta-H° / R times (1/T2 - 1/T1). For an exothermic reaction (Delta-H° less than 0) K decreases as T rises. For an endothermic reaction K increases. This is the only way to actually change K - everything else just shifts the position.
At equilibrium the reactants vastly outnumber the products. For water autoionization (Kw = 10^-14) the equilibrium H+ concentration in pure water is 10^-7 M - one part in 555 million water molecules has dissociated. Tiny K values mean the forward direction is thermodynamically unfavorable.
By Delta-G° = -RT ln K, so K = exp(-Delta-G° / RT). A 5.7 kJ/mol change in Delta-G° corresponds to one order of magnitude in K at 298 K. Knowing one of the pair gives the other immediately.
ICE stands for Initial - Change - Equilibrium. You write the starting concentrations, the unknown change x as a function of stoichiometry (-a x for reactants, +c x for products), and the equilibrium values. Plug them into the K expression and solve for x. The calculator on this page assumes you have already done the ICE step and have equilibrium concentrations.
Q is calculated with the same expression as K but using whatever concentrations the system currently has, not necessarily equilibrium values. Comparing Q to K tells you which way the reaction will shift: Q smaller than K shifts forward, Q larger than K shifts back, Q equal to K means no net change.
No - K is always positive because it is a ratio of concentrations raised to positive powers. ln K can be negative (when K is between 0 and 1) and so can Delta-G°. The signs of those derived quantities tell you which side is favored.