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Kp Calculator (Equilibrium Constant)

Ready to calculate
Two Modes.
13 Pressure Units.
Auto Δn Tracking.
100% Free.
No Data Stored.

How it Works

01Pick a Mode

Convert Kp ↔ Kc, or compute Kp from partial pressures

02Enter Inputs

Mode 1: Kc, T, mol products, mol reactants. Mode 2: 4 partials × 4 coefficients.

03Apply the Formula

Mode 1: Kp = Kc × (RT)^Δn. Mode 2: Kp = (PC^c·PD^d) / (PA^a·PB^b).

04Read the Result

Kp value, Δn, breakdown, and Kp in all common pressure units

What is the Kp Calculator?

The Kp Calculator handles the two most common equilibrium-constant tasks in gas-phase chemistry. Mode 1 — Convert between Kp and Kc: use the relationship Kp = Kc · (RT)^Δn, where Δn is the change in moles of gas going from reactants to products. Mode 2 — Compute Kp directly from partial pressures: for the general equilibrium aA(g) + bB(g) ⇌ cC(g) + dD(g), apply Kp = (P_C^c · P_D^d) / (P_A^a · P_B^b). Both modes are essential for thermodynamics problems, reaction yield analysis, and engineering equilibrium calculations.

Kc (concentration-based) and Kp (pressure-based) are equivalent ways to express the same equilibrium — just measured in different units. They're related by the ideal gas law: PV = nRT means concentration (n/V) and pressure are proportional at constant T. The Δn factor accounts for how the unit conversion compounds when more or fewer moles of gas appear on the product side.

Built for chemistry students learning equilibrium, physical chemistry researchers, chemical engineers calculating reactor yields, and instructors. Free, fast, mobile-friendly, fully client-side.

Pro Tip: When Δn = 0 (equal moles of gas on both sides, e.g., H₂ + I₂ ⇌ 2HI), Kp = Kc regardless of temperature — the (RT)^Δn factor becomes 1.

How to Use the Kp Calculator?

Pick a Mode: "Convert between Kp and Kc" if you have Kc and need Kp; "Calculate Kp from partial pressures" if you have measured pressures.
Convert Mode — Inputs: Kc value, target Kp pressure unit (atm/kPa/bar/Torr/mmHg), temperature, total moles of gaseous products, total moles of gaseous reactants.
Partial Pressure Mode — Inputs: Partial pressures of A, B, C, D (with shared unit dropdown — 13 pressure units supported) and the stoichiometric coefficients a, b, c, d.
Press Calculate: The tool applies the right formula automatically and computes Δn behind the scenes.
Read the Result: Kp value, Δn, full step-by-step breakdown, and (in convert mode) Kp shown in all 5 common pressure units simultaneously.

How is Kp calculated?

The two core equations: Kp = Kc · (RT)^Δn for the conversion, and Kp = ∏ P_products^coeff / ∏ P_reactants^coeff for the direct calculation. R = 0.08206 L·atm/(mol·K) when Kp is expressed in atm.

For Kp expressed in other units (kPa, bar, Torr, mmHg), the calculator multiplies the atm-result by the appropriate unit-conversion factor raised to the Δn power. Internally, all conversions go through atm as the intermediate.

Math — Step by Step:

1. Determine Δn

Δn = sum of gas product moles − sum of gas reactant moles:

  • For aA + bB ⇌ cC + dD: Δn = (c + d) − (a + b)
  • Only gas-phase species count
  • Solids and liquids are excluded

Example: N₂ + 3H₂ ⇌ 2NH₃ → Δn = 2 − (1 + 3) = −2.

2. Apply Kp = Kc·(RT)^Δn

For Kp in atm:

  • R = 0.08206 L·atm/(mol·K)
  • T in kelvins
  • Result in atm^Δn

Example: Kc = 4 at 500 K, Δn = +1 → Kp = 4 × (0.08206 × 500)^1 = 164 atm.

3. Direct from Partial Pressures

For aA + bB ⇌ cC + dD:

  • Kp = (P_C^c · P_D^d) / (P_A^a · P_B^b)
  • All pressures in the same unit
  • Result has units of P^Δn

If a = b = c = d = 1 and pressures equal: Kp = 1.

4. Interpretation

What Kp tells you:

  • Kp > 1: products favored (forward reaction proceeds)
  • Kp ≈ 1: balanced equilibrium
  • Kp < 1: reactants favored (reverse direction proceeds)
  • Kp >> 1 or << 1: reaction goes ~completely

Kp depends only on temperature for a given reaction — independent of starting concentrations.

When Δn Matters:

Δn = 0

Kp = Kc

Same number of moles on both sides. Example: H₂(g) + I₂(g) ⇌ 2HI(g). Temperature change doesn't shift the Kp/Kc ratio.

Δn > 0

Kp > Kc

More gas products. Example: N₂O₄(g) ⇌ 2NO₂(g), Δn = +1. (RT)^Δn multiplier amplifies Kc.

Δn < 0

Kp < Kc

More gas reactants. Example: Haber: N₂ + 3H₂ ⇌ 2NH₃, Δn = −2. (RT)^Δn shrinks Kc.

Real-World Example

Real Equilibrium Examples

Real equilibria — Kp values and Δn:

Reaction T (K) Δn Kc Kp (atm)
H₂ + I₂ ⇌ 2HI500 K0160160 (= Kc)
N₂O₄ ⇌ 2NO₂298 K+14.6e-30.113 atm
N₂ + 3H₂ ⇌ 2NH₃ (Haber)700 K−20.651.97e-4 atm⁻²
2SO₂ + O₂ ⇌ 2SO₃700 K−13.50.0609 atm⁻¹
PCl₅ ⇌ PCl₃ + Cl₂523 K+10.0411.76 atm
CO + Cl₂ ⇌ COCl₂450 K−15.0e3135 atm⁻¹

Notice how Δn determines the units of Kp (atm^Δn) and whether Kp is bigger or smaller than Kc. The Haber process Δn = −2 makes Kp dramatically smaller than Kc at high T.

Who Should Use the Kp Calculator?

1
🧪 Chemistry Students: Convert between Kp and Kc fluently — common exam question type.
2
🔬 Physical Chemistry Researchers: Compute equilibrium constants from gas-phase reactions for thermodynamic analysis.
3
🏭 Chemical Engineers: Reactor design — Kp determines theoretical yield at operating temperature and pressure.
4
🎓 Educators: Generate problem sets, demonstrate Δn dependence, illustrate equilibrium dynamics.
5
⚗️ Catalysis Researchers: Compare Kp at different temperatures to reveal activation thermodynamics (Van't Hoff plots).
6
📚 Industrial Chemistry: Haber-Bosch, Contact process, water-gas shift — all rely on knowing Kp at process conditions.

Technical Reference

Key Takeaways

Kp is the most natural equilibrium constant for gas-phase reactions because gases are easier to measure by pressure than concentration. Use the ToolsACE Kp Calculator to switch between Kp and Kc using the textbook formula Kp = Kc·(RT)^Δn, or compute Kp directly from measured partial pressures. The dual-mode design covers the two most common types of equilibrium-constant problems in undergraduate physical chemistry and chemical engineering. Always remember Δn — it determines whether Kp is bigger or smaller than Kc and what units Kp carries.

Frequently Asked Questions

What is Kp?
Kp is the equilibrium constant for a gas-phase reaction expressed in terms of partial pressures. For the general reaction aA(g) + bB(g) ⇌ cC(g) + dD(g): Kp = (P_C^c · P_D^d) / (P_A^a · P_B^b). It tells you the ratio of product to reactant partial pressures at equilibrium.
What's the difference between Kp and Kc?
Kp uses partial pressures (in atm, kPa, bar, etc.). Kc uses concentrations (in mol/L). Both describe the same equilibrium but in different units. They're related by Kp = Kc·(RT)^Δn, where Δn = (sum of gas product moles) − (sum of gas reactant moles).
What is Δn?
Δn is the change in moles of gas going from reactants to products: Δn = (c + d) − (a + b) for aA + bB ⇌ cC + dD where all are gases. It only counts gas-phase species — solids and liquids are excluded. Δn determines whether Kp is bigger or smaller than Kc and what units Kp has.
What value of R should I use?
For Kp in atm, use R = 0.08206 L·atm/(mol·K). For Kp in kPa, use R = 8.314 L·kPa/(mol·K). The calculator uses R = 0.08206 internally and converts to other Kp units afterward, so just pick your desired Kp unit.
Why does temperature matter so much?
Because the (RT)^Δn factor in Kp = Kc·(RT)^Δn is highly temperature-dependent. At higher T, the term grows. Kp itself also varies with T according to the Van't Hoff equation: lnK = −ΔH/(RT) + ΔS/R. For exothermic reactions, K decreases with T; for endothermic, K increases.
When does Kp = Kc?
When Δn = 0, i.e., the same number of gas-phase moles on both sides of the equation. Examples: H₂(g) + I₂(g) ⇌ 2HI(g); N₂(g) + O₂(g) ⇌ 2NO(g); CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g). For these, Kp and Kc are numerically identical at all temperatures.
What does it mean if Kp = 1?
Kp = 1 indicates a balanced equilibrium where the partial-pressure quotient at equilibrium equals 1. Neither products nor reactants are strongly favored — the reaction reaches equilibrium with comparable amounts of each. In a Δn = 0 reaction this means equal partial pressures; in a Δn ≠ 0 reaction it means a specific weighted balance.
Are solid and liquid reactants included in Kp?
No — only gas-phase species appear in Kp. Pure solids and liquids have constant activity (= 1) and don't show up in the equilibrium expression. Same goes for water in dilute aqueous reactions. For mixed-phase reactions, only the gas pressures appear in Kp.
What units does Kp have?
Kp has units of P^Δn, where P is the chosen pressure unit. Examples: for the Haber reaction (Δn = −2), Kp has units atm⁻². For 2NO₂ ⇌ N₂O₄ (Δn = −1), Kp has units atm⁻¹. When Δn = 0, Kp is dimensionless. Many textbooks suppress the units, treating Kp as a pure number.
What's the difference between Kp and the reaction quotient Qp?
Kp is the equilibrium constant — fixed at any given temperature. Qp uses the same formula but with current (non-equilibrium) partial pressures. Comparing Qp to Kp tells you which direction the reaction will shift: Qp < Kp → forward; Qp > Kp → reverse; Qp = Kp → at equilibrium.
Can I use this for non-ideal gases?
For accurate work at high pressure or near-condensation conditions, replace partial pressures with fugacities (effective pressures) in the Kp expression. The calculator assumes ideal gas behavior — fine for most introductory problems and standard lab conditions, but not for industrial high-pressure equilibria like Haber-Bosch at full operating P.
How does Le Chatelier's principle relate?
Le Chatelier's principle describes how an equilibrium shifts when stressed — by adding/removing reactants or products, changing T, or changing P. Kp itself only changes with temperature (it's a function of T alone for a given reaction). Pressure changes shift the position of equilibrium (Q changes) but not Kp itself.
Is my data private?
All calculations happen locally in your browser. Nothing is sent to a server, saved, or logged. The tool is free and requires no sign-up.

Author Spotlight

The ToolsACE Team - ToolsACE.io Team

The ToolsACE Team

Our chemistry tools team implements both standard Kp calculations: the Kc → Kp conversion via Kp = Kc·(RT)^Δn (using R = 0.08206 L·atm/(mol·K)), and the direct partial-pressure form Kp = (PC^c·PD^d)/(PA^a·PB^b) for the general gas-phase equilibrium aA + bB ⇌ cC + dD.

Chemical EquilibriumIdeal Gas LawSoftware Engineering Team

Disclaimer

Kp assumes ideal gas behavior — accurate at low to moderate pressures. At high pressure or near-condensation conditions, real-gas corrections (fugacities, compressibility factors) become necessary.