The equilibrium constant calculator above takes the balanced coefficients of a reaction and the concentrations or partial pressures measured once the system has stopped changing, and returns the equilibrium constant. It raises each species to the power of its stoichiometric coefficient, divides the product term by the reactant term, and reports the answer alongside the mole change Δn, the converted constant on the other scale, and the standard free energy change that the constant implies.
Arb Digital publishes free calculators for people who need one specific number without a sign-up or a paywall. This one exists because the equilibrium expression is easy to write down and easy to get wrong: a coefficient becomes a multiplier instead of an exponent, a solid ends up in the denominator, or Kp and Kc get used interchangeably when the reaction changes the number of gas molecules. The tool lays out each term so you can see where every number went.
What This Equilibrium Constant Calculator Does
Give it up to two reactants and two products, each with a coefficient and an equilibrium amount, and it builds the mass-action expression for you. The headline result is the equilibrium constant on whichever scale your inputs used. The supporting grid shows the constant converted to the other scale, the value of Δn, the standard Gibbs free energy change calculated from the constant, and a plain-language reading of whether products or reactants dominate.
Below the grid, a set of bars shows the numerator and denominator of the expression side by side. That view answers the question students actually have, which is not what K equals but why it came out so large or so small. When one bar is a sliver next to the other, the reason is visible rather than arithmetic.
Two boundaries are worth stating. This page assumes you already have equilibrium amounts. If you are working from a balanced equation and need the coefficients first, the chemical equation balancer does that job. If you have a constant already and want the free energy consequence, the Gibbs free energy calculator handles ΔG = ΔH − TΔS in full, including the enthalpy term this page never sees.
How to Use It
- Balance the equation first. The coefficients you type are the exponents in the expression, so an unbalanced equation gives a meaningless constant.
- Enter coefficients and equilibrium amounts for each species. Set both boxes to zero for any slot you do not need.
- Leave out pure solids and pure liquids. Their activity is defined as 1, so they contribute nothing to the expression. A solid in the equation still gets a coefficient of 0 here.
- Choose the input scale. Concentrations in mol/L give Kc; partial pressures in atm give Kp. The tool converts to the other one automatically.
- Set the temperature in kelvin. Both the Kc–Kp conversion and the free energy figure depend on it.
The Formula and How It Is Calculated
For the general reaction aA + bB ⇌ cC + dD, the equilibrium constant in terms of concentration is Kc = ([C]c × [D]d) ÷ ([A]a × [B]b). Products go on top, reactants on the bottom, and each coefficient becomes an exponent. The same construction using partial pressures gives Kp.
The link between the two runs through the change in the number of moles of gas: Δn = (c + d) − (a + b), counted over gaseous species only. Then Kp = Kc × (RT)Δn, with R = 0.082057 L·atm·mol⁻¹·K⁻¹ when pressures are in atmospheres and concentrations in mol/L. That value is the CODATA molar gas constant of 8.314462618 J·mol⁻¹·K⁻¹ expressed in the units this conversion needs.
The worked example loaded by default is nitrogen and hydrogen forming ammonia: N₂ + 3H₂ ⇌ 2NH₃. With [N₂] = 0.20 M, [H₂] = 0.40 M and [NH₃] = 0.60 M, the expression gives 0.60² ÷ (0.20 × 0.40³) = 0.36 ÷ 0.0128 = 28.125. Here Δn = 2 − 4 = −2, so at 500 K the conversion gives Kp = 28.125 ÷ (0.082057 × 500)² = 28.125 ÷ 1683.3 = 0.01671. The free energy figure comes from ΔG° = −RT ln K, which for K = 28.125 at 500 K is about −13.9 kJ/mol. The construction of the expression follows the treatment in the OpenStax chapter on equilibrium constants on Chemistry LibreTexts.
When Kc and Kp Are the Same Number, and When They Are Not
The two constants are numerically equal whenever Δn is zero, which happens whenever the reaction has the same number of gas molecules on each side. The hydrogen iodide equilibrium H₂ + I₂ ⇌ 2HI is the textbook case: two gas molecules become two gas molecules, (RT)0 is 1, and the same number serves both scales. That is why so many introductory problems quietly use K without saying which one is meant.
The moment Δn is non-zero the two diverge fast, because RT is a large number. At 500 K, RT in the units used here is about 41. A reaction with Δn = −2 has a Kp roughly 1,700 times smaller than its Kc. A reaction with Δn = +1, such as a decomposition that releases a gas, has a Kp about 41 times larger. Quote a Kp where a Kc was expected in a case like that and the answer is not slightly off, it is off by three orders of magnitude.
Δn counts gases only. A reaction between an aqueous ion and a solid that produces a gas has a Δn of +1 even though three species appear in the equation, because the aqueous species and the solid are not gases. Getting Δn wrong is the single most common failure in this conversion, and it fails silently: the arithmetic still runs and still produces a plausible-looking number.
Why the Constant Has No Units in Serious Work
Write out the expression for ammonia synthesis and the units do not cancel: M² over M⁴ leaves M⁻². Textbooks then either attach those units or quietly drop them, and students reasonably wonder which is right. The formal answer is that an equilibrium constant is built from activities, not concentrations, and an activity is a ratio of a species' effective concentration to its value in a defined standard state. That ratio is dimensionless, so the constant is dimensionless too.
This is also why the ΔG° figure in the grid deserves a caveat. It uses the constant on the scale you entered, and free energy is defined against a specific standard state. If your inputs were concentrations, that number is the free energy change for the solution standard state; if they were pressures in atmospheres, it approximates the gas-phase value on the 1 bar standard state. For a homework answer this distinction never bites. For a published thermodynamic quantity it does.
Reading the Size of K Without Doing Anything Else
The magnitude of K tells you where the equilibrium sits, and the thresholds worth memorising are cruder than most people expect. A K above about 10³ means the reaction goes essentially to completion; at equilibrium there is so little reactant left that ignoring it introduces less error than your measurement does. A K below about 10⁻³ means the reverse: the reaction barely happens, and treating the initial concentrations as the equilibrium concentrations is a good approximation.
Between those bounds you have a genuine mixture, and this is the region where an ICE table earns its keep and where approximations break. The often-taught five percent rule, which says you can neglect x relative to the initial concentration if x is under five percent of it, is a rule about this middle band. It fails exactly when K sits near the initial concentration in magnitude, and it fails without warning.
The Reaction Quotient Is the Same Expression at the Wrong Time
Q, the reaction quotient, uses precisely the formula on this page but with concentrations taken at any moment rather than at equilibrium. That makes the comparison between Q and K the most useful single test in equilibrium work. If Q is less than K the system has too much reactant and will shift forward. If Q is greater than K it has too much product and will shift back. If they are equal it is already at equilibrium and nothing net happens.
You can use this calculator to compute Q by entering non-equilibrium amounts, because the arithmetic is identical. What changes is the interpretation, and the tool cannot tell the difference. If the amounts you typed were not measured at equilibrium, the number on the screen is Q, not K, and it will change as the reaction proceeds. This is worth being deliberate about, because a value obtained from a half-finished reaction and then quoted as a constant is a mistake that survives all the way to a wrong conclusion.
Le Chatelier's principle is really a qualitative shorthand for the Q-versus-K comparison. Adding reactant lowers Q below K, so the system moves forward. Compressing a gas-phase equilibrium raises all partial pressures, and whether Q rises or falls relative to K depends entirely on the sign of Δn — which is why compression shifts ammonia synthesis toward product and does nothing at all to the hydrogen iodide equilibrium.
Where Temperature Actually Enters
Temperature is the only variable that changes K itself. Concentration changes, pressure changes and catalysts all shift the position of the equilibrium without touching the constant. Heat a reaction and the constant genuinely moves, in a direction set by the sign of ΔH through the van 't Hoff relation. Endothermic reactions have a K that rises with temperature; exothermic reactions have a K that falls.
That has a practical consequence for the ammonia example. The reaction is exothermic, so a high temperature lowers the constant and reduces the yield — yet industrial synthesis runs hot, because at low temperature the rate is hopeless. The compromise between a favourable constant and a workable rate is the whole engineering problem, and it is why a catalyst matters so much: it buys back rate without paying for it in equilibrium position. If you are working with rate constants and temperature rather than equilibrium, the half-life calculator covers first-order decay.
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Browse All Free Tools Suggest a ToolCommon Mistakes to Avoid
- Multiplying by the coefficient instead of raising to it — 2 × 0.60 is 1.2, but 0.60² is 0.36. This single slip changes the answer by more than a factor of three.
- Including solids or pure liquids — solid calcium carbonate and liquid water in a heterogeneous equilibrium have an activity of 1 and never appear in the expression.
- Counting non-gases in Δn — Δn is the change in moles of gas only. Aqueous species and solids are excluded even though they sit in the equation.
- Quoting K without its temperature — the constant is only constant at fixed temperature, so a value with no temperature attached cannot be checked or reused.
- Reading a large K as a fast reaction — magnitude describes the destination, not the journey. Kinetics and thermodynamics answer different questions.
Related Free Tools From Arb Digital
Get your coefficients right first with the chemical equation balancer, then turn the constant into a free energy picture with the Gibbs free energy calculator or derive the entropy term with the entropy change calculator. For gas-phase work, the ideal gas law calculator converts between amount, pressure and volume. Acid and base equilibria are handled directly by the pH calculator, and solution preparation by the molarity calculator. The full free online tools hub lists everything else.
Frequently Asked Questions
It is the ratio of product amounts to reactant amounts at equilibrium, with each species raised to the power of its stoichiometric coefficient. For aA + bB reacting to cC + dD, K equals the product terms multiplied together over the reactant terms multiplied together.
Multiply Kc by (RT) raised to the power Δn, where Δn is the change in the number of moles of gas from reactants to products and R is 0.082057 L·atm per mol per kelvin when pressures are in atmospheres. When Δn is zero the two constants are equal.
Their activity is defined as 1 because a pure condensed phase has a fixed effective concentration that does not change as the reaction proceeds. Adding more solid calcium carbonate to a heterogeneous equilibrium does not shift it for exactly this reason.
Formally no. It is built from activities, which are ratios against a standard state and therefore dimensionless. Units appear only when concentrations are substituted directly for activities, which is a shortcut that works because the standard state is 1 mol/L.
That the equilibrium lies far to the product side, so the reaction goes close to completion. Above about a thousand, the leftover reactant is usually smaller than the measurement error. It says nothing about how quickly the reaction gets there.
They are the same expression evaluated at different times. Q uses concentrations at any moment, K uses them at equilibrium. Comparing the two predicts the direction of change: Q below K shifts forward, Q above K shifts backward.
No. A catalyst lowers the activation barrier in both directions equally, so equilibrium is reached sooner but at the same position. Only a change in temperature changes the value of K itself.
This calculator is provided for education and general reference. It describes how an equilibrium constant is computed and is not laboratory or safety guidance; follow the procedures and risk assessments issued by your own institution.
