About the Equilibrium Constant Calculator
This equilibrium constant calculator evaluates the mass-action expression for a general reaction aA + bB ⇌ cC + dD. Enter the stoichiometric coefficients and the equilibrium molar concentrations (for Kc) or partial pressures in atm (for Kp), and it returns the equilibrium constant, its base-10 logarithm, and the matching constant on the other basis using Kp = Kc(RT)^Δn.
It is built for general-chemistry students checking homework, for teachers preparing worked solutions, and for anyone who needs a quick Kc-to-Kp conversion at a given temperature. If you enter non-equilibrium amounts, the same expression gives the reaction quotient Q; add a known K and the calculator tells you which way the reaction will shift to reach equilibrium.
Pure solids and pure liquids do not appear in an equilibrium expression — set their coefficient to 0 so they are skipped. The calculation assumes ideal behaviour (activities equal to molarity or pressure in atm), which is the convention used in introductory and most AP/IB courses.
With the default inputs, the k (or q) from your values is 60.84. Change any value above to recalculate instantly.
How to use the equilibrium constant calculator
- 1Write the balanced equation and map species to A, B (reactants) and C, D (products).
- 2Choose whether you are entering concentrations (Kc) or partial pressures (Kp).
- 3Enter each coefficient and value; set the coefficient to 0 for unused slots, solids and liquids.
- 4Enter the temperature if you want the Kc ↔ Kp conversion.
- 5Optionally enter a known K to see whether the reaction shifts left or right.
Formula and method
The law of mass action says that at equilibrium the ratio of product activities to reactant activities, each raised to its stoichiometric coefficient, is a constant at a given temperature. Using molar concentrations gives Kc; using partial pressures in atmospheres gives Kp. If the values are not at equilibrium, the identical expression is called the reaction quotient Q.
Kc and Kp are linked through the ideal gas law (P = cRT): Kp = Kc(RT)^Δn, where R = 0.082057 L·atm/(mol·K), T is the absolute temperature and Δn is the change in moles of gas. Coefficients set to 0 are excluded, which is how pure solids and liquids are handled. Comparing Q with K predicts the direction of change: Q < K moves right, Q > K moves left.
- [X] or P(X)
- Equilibrium molarity (mol/L) or partial pressure (atm) of species X
- a, b, c, d
- Stoichiometric coefficients of the balanced equation
- R
- Gas constant, 0.082057 L·atm/(mol·K)
- T
- Absolute temperature (K)
- Δn
- Moles of gaseous products − moles of gaseous reactants
Worked examples
H₂ + I₂ ⇌ 2HI at 698 K
With [H₂] = [I₂] = 0.10 M and [HI] = 0.78 M, Kc = 0.78² ÷ (0.10 × 0.10) = 60.84. Because two moles of gas form two moles of gas, Δn = 0 and Kp equals Kc.
Haber process, N₂ + 3H₂ ⇌ 2NH₃ at 773 K
Kc = 0.157² ÷ (0.921 × 0.763³) ≈ 0.0603. Four moles of gas become two, so Δn = −2 and Kp = 0.0603 × (0.082057 × 773)⁻² ≈ 1.50 × 10⁻⁵.
N₂O₄ ⇌ 2NO₂ from partial pressures at 298 K
Kp = 0.6² ÷ 0.4 = 0.90. With Δn = +1, Kc = Kp ÷ (RT) = 0.90 ÷ (0.082057 × 298) ≈ 0.0368.
Is the HI mixture at equilibrium if K = 50?
The same mixture gives Q = 60.84. Since Q is larger than K = 50, there are too many products, so some HI decomposes and the reaction shifts toward H₂ and I₂.
Frequently asked questions
What is the difference between Kc and Kp?+
Kc is written with molar concentrations (mol/L) and Kp with partial pressures (usually atm). They describe the same equilibrium and are related by Kp = Kc(RT)^Δn, so they are equal only when the number of gas moles does not change.
Why are solids and liquids left out of the equilibrium expression?+
The activity of a pure solid or pure liquid is defined as 1, because its concentration does not change during the reaction. Including it would just multiply by 1, so it is omitted — set its coefficient to 0 here.
What is the reaction quotient Q?+
Q has exactly the same form as K but uses the current, not equilibrium, concentrations or pressures. If Q < K the reaction proceeds forward; if Q > K it runs in reverse; if Q = K the system is already at equilibrium.
Does K have units?+
Thermodynamic equilibrium constants are dimensionless because they use activities relative to a standard state. In many textbooks Kc and Kp are written with units such as M² or atm⁻¹, but the numerical value is the same either way.
Does the equilibrium constant change with concentration?+
No. Changing concentrations shifts the position of equilibrium but not K. Only temperature changes K, as described by the van ’t Hoff equation.
What does a large or small K mean?+
A K much greater than 1 means products dominate at equilibrium; a K much less than 1 means reactants dominate. Values between roughly 10⁻³ and 10³ indicate appreciable amounts of both.