From the initial state and the constant K, find the equilibrium.

Kc uses molar concentrations of aqueous and gaseous species (mol/L).

Reaction and coefficients

\mathrm{A}_{(\text{aq})} + \mathrm{B}_{(\text{aq})} \rightleftharpoons \mathrm{C}_{(\text{aq})} + \mathrm{D}_{(\text{aq})}

Adjust each coefficient and phase. Kc includes gases (g) and aqueous species (aq); Kp includes gases only. Pure solids (s) and liquids (l) are excluded. A coefficient of 0 disables the species.

Initial state

Reactant A

Initial concentration

mol/L

Reactant B

Initial concentration

mol/L

Product C

Initial concentration

mol/L

Product D

Initial concentration

mol/L

Equilibrium constant

Numerical value of K under the selected concentration or pressure convention.

1.000e-61.000e+6

Results

Shift toward products (→)

Extent of reaction

0.5

mol/L

Q₀ = 0 | K = 1

Limiting-reactant conversion

Q at equilibrium = 1

Approximation invalid (change ≥ 5%)

ICE Table

ICE Table (mol/L)

Initial concentrations or pressures, stoichiometric changes, and equilibrium values for each species
Species	A(aq)	B(aq)	C(aq)	D(aq)
Initial (I)	1	1	0	0
Change (C)	−0.5	−0.5	+0.5	+0.5
Equilibrium (E)	0.5	0.5	0.5	0.5

Fundamentals & Explanation

What is an ICE table?

When a reversible reaction $aA + bB \rightleftharpoons cC + dD$ reaches equilibrium, the concentrations stop changing but are rarely the initial ones. The ICE table (Initial, Change, Equilibrium) is the standard method to find those equilibrium concentrations: it lays out, in three rows, the Initial state, the Change imposed by stoichiometry, and the Equilibrium state.

The entire change is written with a single unknown, the extent of reaction $\;x$ . Since each species reacts in proportion to its coefficient, the equilibrium concentrations are:

[A]_{eq} = [A]_0 - a\,x \qquad [B]_{eq} = [B]_0 - b\,x

[C]_{eq} = [C]_0 + c\,x \qquad [D]_{eq} = [D]_0 + d\,x

Here $x$ is the ICE-table change variable, not the extensive thermodynamic extent $\xi$ measured in moles. At constant volume, $x_c=\xi/V$ for Kc; for an ideal gas at constant temperature and volume, $x_p=RT\xi/V$ for Kp. This is why the interface expresses $x$ in mol/L or atm.

The quotient Q sets the direction

Before solving, it helps to know which way the system will move. The reaction quotient $Q$ has the same form as $K$ , but is evaluated at the current concentrations:

Q = \frac{[C]^{c}\,[D]^{d}}{[A]^{a}\,[B]^{b}}

Comparing the initial $Q_0$ with $K$ :

$Q_0 < K$ : too few products → the reaction proceeds to the right ( $x > 0$ ).
$Q_0 > K$ : too many products → the reaction shifts to the left ( $x < 0$ ).
$Q_0 = K$ : the system is already at equilibrium ( $x = 0$ ).

Solving for x

At equilibrium $Q = K$ . Substituting the equilibrium concentrations gives an equation in $x$ :

\frac{([C]_0 + c\,x)^{c}\,([D]_0 + d\,x)^{d}}{([A]_0 - a\,x)^{a}\,([B]_0 - b\,x)^{b}} = K

For $1{:}1$ coefficients this often reduces to a quadratic you can solve by hand; other stoichiometries produce a nonlinear equation. The calculator solves it numerically by bisection in a logarithmic coordinate, without applying the small-change approximation.

Within the ideal single-reaction model, $x$ is bounded by non-negativity. Over that interval $Q$ increases strictly monotonically, so a finite positive K has one representable solution or, when no physical movement is possible, a blocked state.

Heterogeneous equilibria: solids and liquids

When species are in different phases, pure solids and pure liquids(and the solvent in dilute solution) do not appear in K. The reason: their activity is 1 and constant — their amount does not shift the equilibrium as long as some of that phase is present.

\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \qquad K_p = P_{\text{CO}_2}

That is why you pick the phase of each species in the calculator. Kc includes gases $(g)$ and aqueous species $(aq)$ ; Kp includes gases only. Pure solids $(s)$ and liquids $(l)$ are excluded (they stay in the ICE table, but not in the K expression).

The module assumes that every selected pure phase is present in a non-limiting amountthroughout the shift. Because its initial amount is not requested, the result no longer applies if a pure solid or liquid is depleted.

The 5% rule

When $K$ is very small, $x$ is tiny compared with the initial concentrations, and many textbooks approximate $[A]_{eq} \approx [A]_0$ to simplify the algebra. The approximation is considered valid if the relative change of the limiting reactant is below 5%:

\frac{a\,x}{[A]_0} \times 100\% < 5\%

The calculator numerically solves the complete model equation and also reports whether that particular small-change approximation would have been acceptable. The badge does not by itself validate ideality, the selected reaction, or any other chemical assumption.

The method, step by step

Write the balanced reaction.
Tabulate the initial concentrations (row I).
Express the change with $x$ and the coefficients (row C).
Add to get the equilibrium values (row E).
Substitute into the expression for $K$ and solve for $x$ .
Compute the concentrations and check the 5% rule.

And the other way around ("Find K" mode): since $x$ is the only unknown, measuring a single species at equilibrium is enough. That value fixes $x$ , you complete the whole table and obtain $K = Q$ at equilibrium — without knowing every concentration in advance. To obtain a finite positive K, every value included in the K expression must be strictly greater than zero.

Model scope

One balanced reversible reaction, up to two reactants and two products, and one K at fixed temperature; an ideal dilute solution for Kc or ideal gases at constant T and V for Kp. The model does not cover non-ideal activities or fugacities, coupled reactions, volume changes during equilibration, or depletion of pure phases.

Kc vs Kp

Kc uses concentrations of aqueous and gaseous species; Kp uses gas partial pressures only. This module follows the general-chemistry convention of working with numerical values in mol/L and atm. A rigorous thermodynamic constant uses dimensionless activities and a standard pressure of 1 bar. For ideal gases and consistent units, the textbook relation is $\;K_p = K_c\,(RT)^{\Delta n_g}$ , where $\Delta n_g$ counts gas coefficients only.

Related calculators

The dissociation of a weak acid ( $K_a$ ) is, at heart, an ICE problem — you will see it in Ionic Dissociation. Once the acid and its conjugate base already coexist, the Buffer (Henderson-Hasselbalch) calculator uses the logarithmic form of that same equilibrium.

Academic References

[1] Chang, R. & Goldsby, K. A. (2015). Chemistry (12th ed.). McGraw-Hill. (Chemical equilibrium, ICE table, 5% rule.)
[2] Atkins, P. & de Paula, J. (2018). Atkins' Physical Chemistry (11th ed.). Oxford University Press. (Equilibrium constant and standard state.)
[3] Brown, T. L., et al. (2017). Chemistry: The Central Science (14th ed.). Pearson. (Reaction quotient and the direction of change.)
[4] IUPAC (2025). Compendium of Chemical Terminology — Gold Book. (Equilibrium constant, extent of reaction, and standard pressure.)

ICE Table — Chemical Equilibrium

Reaction and coefficients

Initial state

Results

ICE Table

Fundamentals & Explanation

What is an ICE table?

The quotient Q sets the direction

Solving for x

Heterogeneous equilibria: solids and liquids

The 5% rule

The method, step by step

Academic References

What is an ICE table?

The quotient Q sets the direction

Solving for x

Heterogeneous equilibria: solids and liquids

The 5% rule

The method, step by step

Academic References