What the equilibrium constant actually tells you
Kc is a single number that fixes the composition a reversible reaction settles at, no matter which side you start from. Mix hydrogen and iodine, or start with pure hydrogen iodide and let it decompose; at the same temperature both mixtures end up obeying the same ratio of concentrations. That is the content of the law of mass action, and it is why one constant is worth more to you than a shelf of individual experiments.
Two things follow immediately. First, Kc tells you where the reaction stops, not how fast it gets there. A reaction with Kc = 1030 may still take geological time if its activation energy is large; that is a question for the Arrhenius equation, not for equilibrium. Second, Kc is a function of temperature only. Changing concentrations, adding a catalyst, or changing the volume shifts the position of equilibrium but leaves the constant alone. Change the temperature and the constant itself moves.
By modern convention Kc is dimensionless. Each concentration in the expression is really a ratio to the 1 mol/L standard state, so the units cancel even when the exponents do not balance. The IUPAC Green Book sets out this activity-based definition; the practical consequence is that you never write units after a K value, and that a K of 4 means the same thing whichever way the exponents fall.
How the expression is built, term by term
Write the balanced equation first, because every exponent comes from it. For aA + bB ⇌ cC + dD the constant is the product concentrations raised to their coefficients, divided by the reactant concentrations raised to theirs. Products go on top; that is a convention, and it is the reason a large K means a product-rich mixture.
The exponents are not decoration. Doubling the coefficient of a species squares its influence. In H2 + I2 ⇌ 2 HI the HI term is [HI]2, so halving the HI concentration cuts the quotient by a factor of four, not two. This is also why a reaction written backwards has constant 1/K, and why a reaction multiplied through by 2 has constant K2.
Leave out pure solids and pure liquids. Their activity is 1 whatever quantity is present, so CaCO3(s) ⇌ CaO(s) + CO2(g) has Kc = [CO2] alone. The same logic makes water disappear from equilibrium expressions in dilute aqueous solution, which is why the ion-product Kw used in the pH calculator contains only [H+] and [OH−].
To go the other way — from Kc and a set of starting amounts to the final composition — you introduce the extent of reaction x, in moles per litre. Every reactant falls by its coefficient times x; every product rises by its coefficient times x. Substituting those expressions into the quotient gives one equation in one unknown. For simple stoichiometry it is a quadratic; for 2 HI or for A + 3B it is a cubic or quartic, and that is where hand algebra becomes painful. This calculator solves it numerically by bisection on ln Q, which is monotonic in x and therefore always has exactly one root in the physically allowed window.
Worked example: H₂ + I₂ ⇌ 2 HI with Kc = 54.3
Charge a vessel with 0.500 M H2 and 0.500 M I2 at 430 °C, where Kc = 54.3, and start with no HI.
- Set up the ICE table. [H2] = 0.500 − x, [I2] = 0.500 − x, [HI] = 0 + 2x.
- Write the quotient. Kc = (2x)2 / [(0.500 − x)(0.500 − x)] = 54.3.
- Exploit the perfect square. Both sides are squares, so take the square root: 2x / (0.500 − x) = √54.3 = 7.36885.
- Clear the fraction. 2x = 3.684427 − 7.36885x, so 9.36885x = 3.684427.
- Solve. x = 3.684427 ÷ 9.36885 = 0.393267 M.
- Back-substitute. [H2] = [I2] = 0.500 − 0.393267 = 0.106733 M; [HI] = 2 × 0.393267 = 0.786535 M.
- Check. (0.786535)2 ÷ (0.106733)2 = 0.618637 ÷ 0.011392 = 54.30. The constant is reproduced, so the arithmetic holds.
Note what the answer means: 78.65% of the hydrogen has been converted. A Kc of 54.3 sounds large, yet more than a fifth of each reactant survives. That gap between "K is big" and "the reaction goes to completion" is where most exam marks are lost.
Now run it backwards to see the other half of the claim. Start instead with 1.000 M HI and nothing else. The same constant gives 2y/(1.000 − 2y) … more directly, [HI]/[H2] must again be 7.36885 at equilibrium, and the calculator returns [HI] = 0.786535 M with [H2] = [I2] = 0.106733 M — the identical mixture, reached from the opposite side.
How to read the number you get
Use magnitude first. A Kc above about 104 means the equilibrium mixture is overwhelmingly product, and modelling the reaction as complete costs you less than a percent. A Kc below about 10−4 means almost nothing happens on its own; if you need the product, you will have to remove it as it forms or couple the step to a downstream reaction. Between those bounds you are in genuine equilibrium territory, where both sides are present in workable amounts and where the ICE table earns its keep.
Read the extent x next. Its sign tells you which way the mixture had to move: positive means the net change ran left to right, negative means products were in excess of what equilibrium allows and the reaction ran backwards. Compare x with the largest value stoichiometry permits — the smallest ratio of a reactant's starting concentration to its coefficient — and you have the fractional conversion. In the worked example the ceiling is 0.500 M and x reached 0.393 M, so conversion is 78.65%.
Finally, sanity-check against thermodynamics. The relation ΔG° = −RT ln K ties the constant to the standard free energy change, so a K of 1 corresponds to ΔG° = 0 and a K of 100 at 298.15 K corresponds to −8.314462618 × 298.15 × ln 100 = −11,416 J/mol, or −11.4 kJ/mol. If a measured K and a tabulated ΔG° disagree badly, one of them is at the wrong temperature. The Gibbs free energy calculator handles that conversion directly.
Equilibrium composition versus Kc for A + B ⇌ 2 C from 0.500 M each
| Kc | √Kc | Extent x (M) | [A] = [B] (M) | [C] (M) | Conversion |
|---|---|---|---|---|---|
| 0.01 | 0.1 | 0.02381 | 0.47619 | 0.04762 | 4.76% |
| 1 | 1 | 0.16667 | 0.33333 | 0.33333 | 33.33% |
| 4 | 2 | 0.25000 | 0.25000 | 0.50000 | 50.00% |
| 54.3 | 7.36885 | 0.39327 | 0.10673 | 0.78653 | 78.65% |
| 100 | 10 | 0.41667 | 0.08333 | 0.83333 | 83.33% |
| 10⁴ | 100 | 0.49020 | 0.00980 | 0.98039 | 98.04% |
| 10⁶ | 1000 | 0.49900 | 0.00100 | 0.99800 | 99.80% |
Conversion is x divided by the 0.500 M ceiling. Notice how slowly conversion approaches 100%: raising Kc from 100 to a million buys only the last 16 percentage points.
Mistakes that produce a wrong Kc
- Using an unbalanced equation. Every exponent comes from a coefficient, so a balancing error propagates straight into the answer. Balance first with the equation balancer.
- Putting initial concentrations into the expression. Kc takes equilibrium values only. Initial values give you Q, not K — a different quantity with a different meaning.
- Including a pure solid or the solvent. Their activities are 1. Writing [CaCO3] or [H2O] into the quotient changes the numerical answer and makes it non-comparable with published values.
- Mixing moles with molarity. The expression needs concentrations. Divide moles by the vessel volume first; the molarity calculator does it in one step.
- Quoting K without a temperature. A constant with no temperature attached cannot be checked, compared, or reused.
- Discarding the wrong root. A quadratic gives two roots and only one keeps every concentration positive. This calculator confines the search to the window where all species stay above zero, so the unphysical root cannot be returned.
- Assuming x is small without checking. The 5% approximation is a shortcut, not a law. If x turns out to exceed 5% of the smallest initial concentration, the approximation has failed and you need the full solution.
What this calculator assumes
It treats concentrations as activities — that is, it assumes an ideal, dilute solution or an ideal gas mixture. In concentrated electrolytes, activity coefficients depart from 1 and a measured constant will drift from the one you compute; analytical chemists handle this with conditional constants at a fixed ionic strength.
It handles up to two reactants and two products. Reactions with more species can still be run by lumping, or by solving the polynomial directly. It also assumes a single equilibrium: coupled equilibria, such as a polyprotic acid or a metal with several ligand complexes, need simultaneous solution rather than one extent variable.
The solver is confined to the window in which every species with a non-zero coefficient stays above zero. If you enter starting amounts where no such window exists — zero on both sides, for instance — it returns a dash rather than a fabricated answer, and the ICE table drops the columns that cannot be filled.
Kc, Kp, Q and the constants that are secretly Kc
For gas-phase reactions the same equilibrium is often written in partial pressures as Kp. The two are related by Kp = Kc(RT)Δn, where Δn is the change in the number of moles of gas; the Kp to Kc conversion calculator handles the arithmetic and the choice of R. When Δn is zero the two constants are numerically identical.
If your mixture is not yet at equilibrium, the same quotient evaluated at the current composition is the reaction quotient Q. Comparing Q with K predicts the direction of net change, which is what the reaction quotient calculator reports.
Several constants you already use are Kc in disguise. Ka and Kb are equilibrium constants for proton transfer, which is why the weak acid pH calculator is an ICE table with a special name. Ksp is the constant for a salt dissolving, with the solid omitted — see the solubility product calculator. Recognising them as one idea saves you learning four sets of rules.
Key terms
- Law of mass action
- The statement that at equilibrium the ratio of product concentrations to reactant concentrations, each raised to its stoichiometric coefficient, is a constant at fixed temperature.
- ICE table
- Initial, Change, Equilibrium — a three-row bookkeeping layout that expresses every equilibrium concentration in terms of one unknown extent.
- Extent of reaction (x)
- How far the reaction has advanced, in moles per litre of reaction as written. Positive means net forward; negative means net reverse.
- Activity
- The effective concentration that appears in a rigorous equilibrium expression: concentration divided by the standard state, corrected by an activity coefficient. Taken as 1 for pure solids and liquids.
- Homogeneous vs heterogeneous equilibrium
- Homogeneous equilibria have all species in one phase; heterogeneous ones involve more than one phase and therefore omit the pure condensed phases from the expression.