Equilibrium Constant Calculator
Calculate the equilibrium constant Kc from concentrations, switch to Kp for gas-phase reactions, convert between the two, and compute the standard Gibbs free energy change. Enter the equilibrium concentrations and stoichiometric coefficients for aA + bB ⇌ cC + dD, choose your mode, and this calculator returns K, log K, delta G, and a comparison of Q vs K.
Formula
Worked example
For 2 A ⇌ B with [B] = 0.5 mol/L and [A] = 0.1 mol/L: K = [B]^1 / [A]^2 = 0.5 / 0.01 = 50. K > 1 so B is favoured. At 298 K, delta-G = -8.314 * 298 * ln(50) / 1000 = -9.69 kJ/mol. For gas-phase work with delta-n = -1: Kp = 50 * (0.08206 * 298)^(-1) = 50 / 24.45 = 2.046.
What the equilibrium constant tells you
When a reversible reaction aA + bB ⇌ cC + dD reaches equilibrium, the forward and reverse rates are equal and the concentrations stop changing. Kc captures that balance as a single dimensionless number: product concentrations raised to their coefficients, divided by reactant concentrations raised to theirs. A large Kc (much greater than 1) means the numerator dominates, so the reaction has gone far toward products before settling. A small Kc (well below 1) means little product forms. Kc near 1 means appreciable amounts of everything coexist. Crucially, Kc depends on the reaction and the temperature alone, not on how much of each species you start with.
Kp and the relationship Kp = Kc(RT)^delta-n
For gas-phase reactions it is often more convenient to express equilibrium in terms of partial pressures rather than concentrations, giving the pressure equilibrium constant Kp. The two constants are linked by Kp = Kc * (RT)^delta-n, where R = 0.08206 L*atm/(mol*K), T is the absolute temperature in kelvin, and delta-n is the difference in total moles of gaseous products minus gaseous reactants. When delta-n = 0, Kp equals Kc exactly. Use the "Convert Kc to Kp" mode here to do this arithmetic automatically. For heterogeneous equilibria, pure solids and liquids are omitted from both K expressions because their activities are taken as 1.
Gibbs free energy and the K relationship
The equilibrium constant is directly linked to thermodynamics through delta-G = -RT ln K, where R is the gas constant 8.314 J/(mol*K) and T is the absolute temperature. A large K gives a large negative delta-G, confirming the forward reaction is thermodynamically spontaneous. This equation also works in reverse: if you know delta-G from a table, you can compute K at any temperature. The Van't Hoff equation extends this further: ln(K2/K1) = -(delta-H/R) * (1/T2 - 1/T1), showing that K increases with temperature for endothermic reactions and decreases for exothermic ones.
The reaction quotient Q and predicting reaction direction
The reaction quotient Q has the same form as K but uses concentrations at any moment, not just at equilibrium. Comparing Q to K predicts the direction the reaction will shift: if Q < K, the system must make more products (forward direction); if Q > K, the system must consume products (reverse direction); if Q = K, the system is already at equilibrium. Switch to the "Reaction quotient Q" mode here to enter your current concentrations, supply the known K, and get an instant direction prediction. This is the quantitative form of Le Chatelier's principle.
Building the expression correctly: what to include
Write the balanced equation first, because the stoichiometric coefficients become the exponents in K. Concentrations are in mol/L for Kc and partial pressures in atm for Kp. Pure solids and pure liquids, including the solvent water in dilute aqueous reactions, are excluded from the K expression because their activity is 1 by convention. In this calculator, simply set the coefficient of any solid or liquid to 0 or leave its fields blank. Always use the equilibrium values, not starting amounts. If you only know initial amounts, build an ICE (Initial, Change, Equilibrium) table to find the equilibrium concentrations first, then bring them here.
Reading K and delta-G together
| K value | log K | delta-G (kJ/mol, 298 K) | Equilibrium position |
|---|---|---|---|
| > 10^6 | > 6 | < -34 | Products essentially complete |
| 10^3 to 10^6 | 3 to 6 | -17 to -34 | Products strongly favoured |
| 10 to 10^3 | 1 to 3 | -6 to -17 | Products moderately favoured |
| 0.1 to 10 | -1 to 1 | -6 to +6 | Comparable amounts (balanced) |
| 10^-3 to 0.1 | -3 to -1 | +6 to +17 | Reactants moderately favoured |
| < 10^-3 | < -3 | > +17 | Reactants strongly favoured |
K and delta-G carry the same information; large K means large negative delta-G at 298 K.
Frequently asked questions
What is the formula for the equilibrium constant K?
For aA + bB ⇌ cC + dD, Kc = ([C]^c * [D]^d) / ([A]^a * [B]^b). Each equilibrium concentration is raised to its stoichiometric coefficient. Products go in the numerator, reactants in the denominator. Concentrations are in mol/L for Kc.
What is the difference between Kc and Kp?
Kc uses molar concentrations (mol/L) and is most convenient for reactions in solution or mixed phases. Kp uses partial pressures (atm) and is the natural choice for gas-phase reactions. They are related by Kp = Kc * (RT)^delta-n, where R = 0.08206 L*atm/(mol*K), T is the temperature in kelvin, and delta-n is the total change in moles of gas (product moles minus reactant moles). When delta-n = 0, Kp and Kc are numerically identical.
How does K relate to Gibbs free energy?
The standard Gibbs free energy change is delta-G = -RT ln K, where R = 8.314 J/(mol*K). A K much greater than 1 gives a large negative delta-G (spontaneous in the forward direction at standard conditions). A K much less than 1 gives a positive delta-G (reverse direction favoured). At K = 1, delta-G = 0, and products and reactants are equally favoured.
What is the reaction quotient Q and how does it differ from K?
Q has the same mathematical form as K but uses any concentrations at a given moment, not the equilibrium values. If Q < K the forward reaction must proceed to reach equilibrium; if Q > K the reverse reaction dominates. K is a fixed number at a given temperature; Q changes as the reaction progresses until it equals K at equilibrium.
Should I include solids and liquids in the K expression?
No. Pure solids and pure liquids, and the solvent water in dilute aqueous reactions, are excluded because their activity is defined as 1 by convention. Only dissolved species and gases appear in K. In this calculator, set the coefficient of any solid or liquid to 0 or leave its fields blank.
Does K change when I add more reactant?
No. K depends only on temperature. Adding reactant shifts the position of equilibrium (the system makes more product to restore the same ratio), but the value of K does not change. Only a temperature change alters K itself.