Equilibrium Constant Calculator — Kc, Kp & ICE Tables
Write Kc and Kp equilibrium constant expressions, calculate Kc from concentrations, solve ICE table problems step-by-step, convert between Kp and Kc using Kp=Kc(RT)^Δn, and predict reaction direction by comparing Q to K — with color-coded ICE tables and full working shown for every calculation.
Build your reaction using the reaction builder below. The equilibrium constant expression (Kc and Kp) will be generated automatically — with pure solids and pure liquids excluded from the equilibrium expression.
Gas-Phase Stoichiometry
Enter the reaction and equilibrium concentrations for each species. The Kc calculator computes the equilibrium constant with full step-by-step working.
Equilibrium Constant
The ICE table method is the universal framework for solving equilibrium problems. Enter your reaction, initial concentrations, and Kc — the ICE table solver finds equilibrium concentrations with full step-by-step working and a color-coded ICE table showing I (Initial), C (Change), and E (Equilibrium) rows.
| Species |
|---|
Equilibrium Concentrations
Convert between Kp (partial pressures) and Kc (molar concentrations) using the relationship Kp = Kc × (RT)^Δn. Enter your reaction to auto-calculate Δn, or enter Δn manually.
Result
The reaction quotient Q has the same mathematical form as Kc but uses current concentrations (not necessarily at equilibrium). Comparing Q to K predicts which direction the reaction will proceed: Q < K means forward, Q > K means reverse, Q = K means equilibrium.
| Species Type | Phase | Include in K? | Reason |
|---|---|---|---|
| Gas | (g) | ✅ YES | Concentration/pressure changes |
| Aqueous solution | (aq) | ✅ YES | Concentration changes |
| Pure solid | (s) | ❌ NO | Constant concentration — absorbed into K |
| Pure liquid | (l) | ❌ NO | Constant concentration — absorbed into K |
| Water (solvent) | (l) | ❌ NO | Pure liquid — constant |
| Water (dilute reactant) | (aq) | ✅ YES | When very dilute, concentration matters |
⚠ Pure solids and pure liquids excluded from all equilibrium constant expressions. Example: CaCO₃(s) ⇌ CaO(s) + CO₂(g) → Kc = [CO₂] only. Both solids are excluded.
| K Value | Products vs Reactants | Reaction Favored | Example |
|---|---|---|---|
| K >> 1 (>10³) | Products greatly favored | Forward ✅ | H₂+½O₂→H₂O, K≈10⁴¹ |
| K > 1 | Products favored | Forward | H₂+I₂⇌2HI, K=55 at 425°C |
| K ≈ 1 | Comparable amounts | Neither | Some reactions at specific T |
| K < 1 | Reactants favored | Reverse | — |
| K << 1 (<10⁻³) | Reactants greatly favored | Reverse ❌ | N₂⇌2N, K≈10⁻⁷⁰ |
| Symbol | Name | Expression | Value at 25°C |
|---|---|---|---|
| Ka | Acid dissociation | [H⁺][A⁻]/[HA] | Varies by acid |
| Kb | Base dissociation | [BH⁺][OH⁻]/[B] | Varies by base |
| Kw | Ion product of water | [H⁺][OH⁻] | 1.0×10⁻¹⁴ |
| Ksp | Solubility product | [M^n+][X^m-]^m | Varies by salt |
| Kp | Pressure equilibrium | Product of partial pressures | = Kc×(RT)^Δn |
| Keq | General equilibrium | Products/reactants | Temperature-dependent |
| Operation | Effect on K | Example |
|---|---|---|
| Reverse a reaction | K_new = 1/K | K=55 → reversed K=1/55=0.018 |
| Multiply coefficients by n | K_new = K^n | K=55, n=2 → K²=3025 |
| Add two reactions | K_overall = K₁ × K₂ | Hess's law analogue |
| Subtract reaction 2 from 1 | K_overall = K₁/K₂ | — |
Technically K is dimensionless — all concentrations are expressed relative to their standard states (c° = 1 M for solutions, p° = 1 atm or 1 bar for gases), making K a pure number without units.
In introductory chemistry, K is treated as having implied units of M^Δn or atm^Δn for calculation purposes, but the numeric values are identical to the dimensionless form. The equilibrium constant does not have units in the rigorous thermodynamic sense.
| Acid | Formula | Ka | pKa |
|---|---|---|---|
| Acetic acid | CH₃COOH | 1.8×10⁻⁵ | 4.74 |
| Formic acid | HCOOH | 1.8×10⁻⁴ | 3.74 |
| Carbonic acid Ka1 | H₂CO₃ | 4.3×10⁻⁷ | 6.37 |
| Carbonic acid Ka2 | HCO₃⁻ | 4.7×10⁻¹¹ | 10.33 |
| Hydrofluoric acid | HF | 7.2×10⁻⁴ | 3.14 |
| Nitrous acid | HNO₂ | 4.5×10⁻⁴ | 3.35 |
| Hydrocyanic acid | HCN | 6.2×10⁻¹⁰ | 9.21 |
| Phenol | C₆H₅OH | 1.0×10⁻¹⁰ | 10.00 |
| Hypochlorous acid | HOCl | 2.9×10⁻⁸ | 7.54 |
| Ammonium ion | NH₄⁺ | 5.6×10⁻¹⁰ | 9.25 |
| Oxalic acid Ka1 | H₂C₂O₄ | 5.9×10⁻² | 1.23 |
| Oxalic acid Ka2 | HC₂O₄⁻ | 6.4×10⁻⁵ | 4.19 |
| Base | Formula | Kb | pKb |
|---|---|---|---|
| Ammonia | NH₃ | 1.8×10⁻⁵ | 4.74 |
| Methylamine | CH₃NH₂ | 4.4×10⁻⁴ | 3.36 |
| Ethylamine | C₂H₅NH₂ | 6.4×10⁻⁴ | 3.19 |
| Pyridine | C₅H₅N | 1.7×10⁻⁹ | 8.77 |
| Aniline | C₆H₅NH₂ | 4.3×10⁻¹⁰ | 9.37 |
| Hydrazine | N₂H₄ | 9.6×10⁻⁷ | 6.02 |
| Reaction | T (K) | Kc | Δn |
|---|---|---|---|
| N₂ + 3H₂ ⇌ 2NH₃ | 298 | 3.7×10⁸ | −2 |
| N₂ + 3H₂ ⇌ 2NH₃ | 773 | 1.7×10⁻⁵ | −2 |
| H₂ + I₂ ⇌ 2HI | 698 | 54.3 | 0 |
| H₂ + I₂ ⇌ 2HI | 425 | 55.3 | 0 |
| 2NO₂ ⇌ N₂O₄ | 298 | 170 | −1 |
| N₂O₄ ⇌ 2NO₂ | 298 | 5.9×10⁻³ | +1 |
| PCl₅ ⇌ PCl₃ + Cl₂ | 250 | 0.0211 | +1 |
| CO + 3H₂ ⇌ CH₄ + H₂O | 500 | 3.92 | −2 |
| CaCO₃ ⇌ CaO + CO₂ | 298 | 3.36×10⁻²⁹ | +1 |
What Is the Equilibrium Constant — Kc, Kp, and Keq
This equilibrium constant calculator computes Kc and Kp from equilibrium concentrations and partial pressures, solves ICE table problems to find equilibrium concentrations from a given Kc, converts between Kp and Kc using Kp=Kc(RT)^Δn, and compares the reaction quotient Q to K to predict reaction direction — with full step-by-step working and color-coded ICE tables for every equilibrium problem.
The equilibrium constant (Kc, Kp, or Keq) is a dimensionless number that quantifies the ratio of product concentrations to reactant concentrations when a reversible reaction reaches chemical equilibrium. For the general reaction aA + bB ⇌ cC + dD:
A large equilibrium constant (K >> 1) means products are strongly favored at equilibrium. A small K (<< 1) means reactants are favored. K ≈ 1 means significant amounts of both reactants and products exist at equilibrium. The equilibrium constant depends only on temperature — not on initial concentrations, total pressure, or the presence of a catalyst.
How to Write an Equilibrium Constant Expression
Writing a correct equilibrium constant expression requires applying four rules consistently. These rules define which species appear in the Kc expression and at what power.
The Four Rules for Equilibrium Expressions
- Products in the numerator, reactants in the denominator — K is products over reactants. This is the universal convention for how to write an equilibrium constant expression.
- Each concentration raised to its stoichiometric coefficient — If 2 mol of HI appear, write [HI]², not 2[HI]. The coefficient becomes the exponent.
- Pure solids (s) and pure liquids (l) are EXCLUDED — their concentrations are constant and absorbed into K. This is the single most common source of error when writing equilibrium constant expressions.
- Gases (g) and aqueous solutions (aq) are INCLUDED — their concentrations change and must appear in the equilibrium expression.
⚠ Pure solids and pure liquids excluded: CaCO₃(s) ⇌ CaO(s) + CO₂(g) → Kc = [CO₂] only. Writing Kc = [CO₂]/[CaCO₃][CaO] is wrong — solids are excluded from all equilibrium constant expressions.
Example 1 — Gas-phase reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
- All species are gases → all included in the equilibrium constant expression
- NH₃ has coefficient 2 → [NH₃]²; H₂ has coefficient 3 → [H₂]³
- Kc = [NH₃]² / ([N₂][H₂]³)
- Kp = (P_NH₃)² / (P_N₂ × P_H₂³)
- Δn = 2 − 4 = −2 → Kp = Kc × (RT)^(−2)
Example 2 — With pure solids excluded: CaCO₃(s) ⇌ CaO(s) + CO₂(g)
- CaCO₃(s) — EXCLUDED from equilibrium expression (pure solid)
- CaO(s) — EXCLUDED from equilibrium expression (pure solid)
- CO₂(g) — INCLUDED (gas)
- Kc = [CO₂] — simplified to just the gas concentration
Example 3 — Aqueous reaction: NH₃(aq) + H₂O(l) ⇌ NH₄⁺(aq) + OH⁻(aq)
- NH₃(aq) — INCLUDED (aqueous)
- H₂O(l) — EXCLUDED from equilibrium expression (pure liquid — solvent)
- NH₄⁺(aq) and OH⁻(aq) — INCLUDED (aqueous)
- Kb = [NH₄⁺][OH⁻] / [NH₃]
The ICE Table Method — Solving Equilibrium Problems
The ICE table method is the universal framework for solving chemical equilibrium problems. ICE stands for Initial (starting concentrations), Change (how concentrations shift as equilibrium is approached), and Equilibrium (final concentrations at equilibrium). The ICE table organizes equilibrium calculations into a three-row table that makes the algebra systematic and error-free.
The ICE table calculator above generates a color-coded ICE table — blue for the Initial row, amber for the Change row showing algebraic expressions (−x, +2x), and green for the Equilibrium row showing final concentrations. This color-coded ICE table makes the structure of every equilibrium problem immediately visible.
How to Set Up an ICE Table — Step by Step
- Write the balanced equation and identify the active species (gases and aqueous — exclude pure solids and pure liquids)
- Fill in the I row with given initial concentrations. Products often start at 0 M.
- Write the C row as algebraic expressions: −coeff×x for reactants (they decrease), +coeff×x for products (they increase)
- Write the E row as E = I + C for each species
- Substitute the E row into the Kc expression and solve for x
- Calculate equilibrium concentrations by substituting x back into the E expressions
- Verify by calculating Kc from the equilibrium concentrations — it should match the given Kc
ICE Table Example: H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 55.3, [H₂]₀ = [I₂]₀ = 1.00 M
- ICE table: I = 1.00, 1.00, 0; C = −x, −x, +2x; E = (1−x), (1−x), 2x
- Kc = (2x)²/(1−x)² = 55.3 → take square root: 2x/(1−x) = 7.436
- 2x = 7.436 − 7.436x → 9.436x = 7.436 → x = 0.7879
- [H₂] = [I₂] = 0.2121 M; [HI] = 1.576 M
- Verify: (1.576)²/(0.2121)² = 2.484/0.0450 = 55.2 ≈ 55.3 ✓
Three Types of ICE Table Algebra
- Perfect square (take square root) — when Kc = (cx)^p/(initial−x)^p, both sides simplify by taking the square root. This is the easiest case — used for H₂ + I₂ ⇌ 2HI and similar symmetric reactions.
- Quadratic formula — when the ICE table produces a quadratic equation ax² + bx + c = 0. Use x = (−b ± √(b²−4ac)) / 2a, taking the physically meaningful (positive) root.
- Small x approximation — valid when Kc is very small (<< 1). Approximate (C₀ − x) ≈ C₀ to simplify the algebra dramatically. Always verify: if x/C₀ > 5%, the approximation is invalid and the quadratic must be used.
The Small x Approximation — 5% Rule
The small x approximation is valid when the change x is negligible compared to the initial concentration. The 5% rule states: if x/[A]₀ × 100% < 5%, the approximation is valid. The small x approximation simplifies ICE table algebra significantly for reactions with small Kc values.
✓ Small x approximation valid example: Ka = 1.8×10⁻⁵, [HA]₀ = 0.100 M → x ≈ √(1.8×10⁻⁵ × 0.100) = 1.34×10⁻³ M. Check: 1.34×10⁻³/0.100 = 1.34% < 5% ✓ Approximation is valid for this ICE table calculation.
✗ Small x approximation invalid: Kc = 0.50, [A]₀ = 0.100 M → x ≈ √(0.50 × 0.100) = 0.224 M which exceeds the initial concentration! Must use the full quadratic formula. The small x approximation fails when Kc is not << 1.
Kp — Equilibrium Constant in Terms of Partial Pressures
For gas-phase reactions, partial pressures can replace molar concentrations in the equilibrium expression. Kp uses partial pressures (in atm or bar); Kc uses molar concentrations (mol/L). Both are valid equilibrium constants for the same reaction at the same temperature — they differ only when Δn ≠ 0.
The Kp/Kc relationship Kp = Kc(RT)^Δn comes from the ideal gas law: since PV = nRT, partial pressure P = (n/V)RT = [concentration] × RT. Substituting into the Kc expression converts each concentration to a pressure, generating the (RT)^Δn factor.
Three Cases for Kp ↔ Kc Conversion
- Δn = 0: Kp = Kc × (RT)⁰ = Kc × 1 = Kc — Kp equals Kc at any temperature. Example: H₂(g) + I₂(g) ⇌ 2HI(g), Δn = 2 − 2 = 0, so Kp = Kc always.
- Δn > 0: Kp > Kc — more moles of gas on the product side. Example: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), Δn = +1.
- Δn < 0: Kp < Kc — fewer moles of gas on the product side. Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn = −2, so Kp < Kc.
Kp Calculation: N₂+3H₂⇌2NH₃, Kc=3.7×10⁸ at T=298.15 K
- Δn = 2 − (1+3) = 2 − 4 = −2
- RT = 0.082057 × 298.15 = 24.465 L·atm/mol
- (RT)^Δn = (24.465)^(−2) = 1/598.5 = 1.671×10⁻³
- Kp = Kc × (RT)^Δn = 3.7×10⁸ × 1.671×10⁻³ = 6.18×10⁵
Q vs K — Predicting Reaction Direction
The reaction quotient Q (also written Qc) has the same mathematical form as the equilibrium constant Kc but uses current concentrations rather than equilibrium concentrations. By comparing Q to K, we can predict which direction a reaction will proceed to reach equilibrium — this is the quantitative basis of Le Chatelier's principle.
The Three Cases: Q < K, Q > K, Q = K
- Q < K → Forward reaction: The system has too many reactants relative to products. The reaction proceeds forward (toward products) until Q = K. More products form.
- Q > K → Reverse reaction: The system has too many products relative to reactants. The reaction proceeds in reverse (toward reactants) until Q = K. More reactants form.
- Q = K → At equilibrium: The system is already at equilibrium. No net reaction occurs in either direction.
Q vs K Example: H₂+I₂⇌2HI, Kc=55.3, current concentrations [H₂]=0.50M, [I₂]=0.50M, [HI]=1.00M
- Q = [HI]²/([H₂][I₂]) = (1.00)²/(0.50×0.50) = 1.00/0.25 = 4.00
- Q = 4.00 < K = 55.3
- Q < K → Forward reaction — the system proceeds toward more HI until Q = K = 55.3
Combining Equilibrium Constants
When reactions are added, subtracted, or scaled, their equilibrium constants combine mathematically. This is the equilibrium analogue of Hess's law for enthalpy changes. Three rules govern how equilibrium constants combine:
- Reverse a reaction: K_new = 1/K — if the forward reaction has K = 55.3, the reverse reaction has K = 1/55.3 = 0.018
- Multiply stoichiometry by n: K_new = K^n — if 2HI ⇌ H₂ + I₂ has K = 0.018, then 4HI ⇌ 2H₂ + 2I₂ has K = (0.018)² = 3.2×10⁻⁴
- Add two reactions: K_overall = K₁ × K₂ — adding equilibrium reactions multiplies their equilibrium constants
Combining K Example: Find K_overall for A ⇌ C given A ⇌ B (K₁=2.5) and B ⇌ C (K₂=4.0)
- Add the reactions: A ⇌ B and B ⇌ C → A ⇌ C (B cancels)
- K_overall = K₁ × K₂ = 2.5 × 4.0 = 10.0
Does the Equilibrium Constant Have Units?
Technically, the equilibrium constant does not have units — K is dimensionless. All concentrations in the Kc expression are divided by the standard state concentration (c° = 1 mol/L), and all pressures in the Kp expression are divided by the standard pressure (p° = 1 atm or 1 bar). This makes every term a pure dimensionless ratio, so Kc and Kp are dimensionless numbers.
In practice, introductory chemistry courses treat K as having implied units of M^Δn (for Kc) or atm^Δn (for Kp) because this makes dimensional analysis of ICE table calculations more straightforward. The numeric values are identical whether or not units are included, so this simplification does not cause computational errors.
📌 Answer: The equilibrium constant Kc and Kp technically have no units (dimensionless). In AP Chemistry and general chemistry courses, K is commonly written without units, and this convention is correct. Does Kc have units? No. Does Kp have units? No. Does Keq have units? No — all equilibrium constants are dimensionless by rigorous thermodynamic definition.
Common Mistakes in Equilibrium Calculations
Mistake 1 — Including pure solids and liquids in K expression
- ❌ Wrong: Kc = [CO₂]/([CaCO₃][CaO]) for CaCO₃(s) ⇌ CaO(s) + CO₂(g)
- ✅ Correct: Kc = [CO₂] — pure solids and pure liquids excluded from equilibrium constant expressions
Mistake 2 — Not raising concentration to the stoichiometric coefficient
- ❌ Wrong: Kc = 2[HI] / ([H₂][I₂]) for H₂ + I₂ ⇌ 2HI
- ✅ Correct: Kc = [HI]² / ([H₂][I₂]) — coefficient 2 becomes the exponent, not a multiplier
Mistake 3 — Using initial concentrations instead of equilibrium concentrations
- ❌ Wrong: substituting initial concentrations directly into Kc formula
- ✅ Correct: use the Equilibrium row of the ICE table — concentrations after the system reaches equilibrium
Mistake 4 — Using small x approximation when it is invalid (>5%)
- ❌ Wrong: using small x approximation when x/[A]₀ = 12% — gives incorrect equilibrium concentrations
- ✅ Correct: always check the 5% rule. If x/[A]₀ × 100% > 5%, use the quadratic formula in the ICE table
Mistake 5 — Wrong sign or calculation for Δn in Kp ↔ Kc conversion
- ❌ Wrong: including solid or liquid species when counting Δn
- ✅ Correct: Δn = moles of gaseous products − moles of gaseous reactants only. For N₂(g) + 3H₂(g) ⇌ 2NH₃(g): Δn = 2 − (1+3) = −2
Equilibrium Practice Problems — 8 Worked Examples
1. Write Kc for N₂ + 3H₂ ⇌ 2NH₃
- Products in numerator, reactants in denominator — all species are gases → all included
- NH₃ has coefficient 2 → [NH₃]²; H₂ has coefficient 3 → [H₂]³
- Kc = [NH₃]² / ([N₂][H₂]³)
2. Calculate Kc for H₂+I₂⇌2HI given [H₂]=0.220 M, [I₂]=0.220 M, [HI]=1.560 M
- Kc = [HI]² / ([H₂][I₂])
- Numerator: (1.560)² = 2.4336
- Denominator: 0.220 × 0.220 = 0.0484
- Kc = 2.4336 / 0.0484 = 50.3
- log(50.3) = +1.70 → products significantly favored
3. ICE table: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g), [PCl₅]₀ = 1.00 M, Kc = 0.0211
- ICE table: I = 1.00, 0, 0; C = −x, +x, +x; E = (1−x), x, x
- Kc = x² / (1−x) = 0.0211
- x² + 0.0211x − 0.0211 = 0 → quadratic formula
- x = (−0.0211 + √(0.000445 + 0.0844)) / 2 = (−0.0211 + 0.2923) / 2 = 0.1356
- [PCl₅] = 0.864 M; [PCl₃] = [Cl₂] = 0.136 M
4. ICE table with square root: H₂+I₂⇌2HI, [H₂]₀=[I₂]₀=1.00 M, Kc=55.3
- ICE table: I = 1.00, 1.00, 0; C = −x, −x, +2x; E = (1−x), (1−x), 2x
- Kc = (2x)²/(1−x)² = 55.3 → √: 2x/(1−x) = 7.436
- x = 0.7879 M
- [H₂] = [I₂] = 0.2121 M; [HI] = 1.576 M
5. Small x approximation: weak acid HA, [HA]₀ = 0.100 M, Ka = 1.8×10⁻⁵
- ICE table: HA ⇌ H⁺ + A⁻; I = 0.100, 0, 0; E = (0.100−x), x, x
- Ka = x²/(0.100−x) ≈ x²/0.100 (small x approximation)
- x² = 1.8×10⁻⁶ → x = 1.34×10⁻³ M
- Check: 1.34×10⁻³/0.100 = 1.34% < 5% ✓ Small x approximation valid
- [H⁺] = [A⁻] = 1.34×10⁻³ M; pH = 2.87
6. Kp from Kc: N₂+3H₂⇌2NH₃, Kc=3.7×10⁸ at T=500 K
- Δn = 2 − 4 = −2 (gas species only)
- RT = 0.082057 × 500 = 41.029 L·atm/mol
- (RT)^(−2) = 1/(41.029)² = 5.93×10⁻⁴
- Kp = 3.7×10⁸ × 5.93×10⁻⁴ = 2.19×10⁵
7. Q vs K: PCl₅⇌PCl₃+Cl₂, Kc=0.0211, current [PCl₅]=0.800 M, [PCl₃]=[Cl₂]=0.050 M
- Q = [PCl₃][Cl₂]/[PCl₅] = (0.050)(0.050)/0.800 = 0.0025/0.800 = 0.00313
- Q = 0.00313 < K = 0.0211
- Q < K → Forward reaction — more PCl₃ and Cl₂ will form
8. Combining K: given A⇌B (K₁=3.0×10²) and 2B⇌C (K₂=1.5×10⁻³), find K for 2A⇌C
- Target: 2A ⇌ C — multiply first reaction by 2 and add second
- 2×(A⇌B): K = (3.0×10²)² = 9.0×10⁴
- K_overall = K₁² × K₂ = 9.0×10⁴ × 1.5×10⁻³ = 135
Frequently Asked Questions
Related Calculators
| Acid | Ka |
|---|---|
| Acetic (CH₃COOH) | 1.8×10⁻⁵ |
| Formic (HCOOH) | 1.8×10⁻⁴ |
| HF | 7.2×10⁻⁴ |
| HNO₂ | 4.5×10⁻⁴ |
| H₂CO₃ (Ka1) | 4.3×10⁻⁷ |
| HCN | 6.2×10⁻¹⁰ |
| NH₄⁺ | 5.6×10⁻¹⁰ |
| HOCl | 2.9×10⁻⁸ |
| Reaction | Kc |
|---|---|
| N₂+3H₂⇌2NH₃ (298K) | 3.7×10⁸ |
| H₂+I₂⇌2HI (698K) | 54.3 |
| 2NO₂⇌N₂O₄ (298K) | 170 |
| PCl₅⇌PCl₃+Cl₂ | 0.0211 |
| N₂O₄⇌2NO₂ (298K) | 5.9×10⁻³ |