This Kp calculator converts between the concentration equilibrium constant Kc and the pressure equilibrium constant Kp for gas reactions: Kp = Kc × (RT)^Δn, where Δn is the change in moles of GAS between products and reactants.

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One Constant, Two Dialects

Equilibrium doesn’t care whether you count gases by concentration or by pressure — but your data source does, and Kp = Kc(RT)^Δn is the translation. The defaults translate the industrially canonical case: ammonia synthesis with Kc = 0.5 at 298 K and Δn = −2 gives Kp = 0.5 × (0.08206 × 298)^(−2) = 8.36 × 10⁻⁴ — the (RT)^Δn factor is a tiny 1.67 × 10⁻³ because two net gas moles VANISH across the reaction, so pressure-based accounting prices products far lower relative to reactants than concentration-based accounting does. The reverse example shows the lever’s other side: Δn = +1 at 500 K multiplies Kc 2.5 by RT = 41.03 into Kp = 102.6. The exponent’s sign and size are pure stoichiometry: count gas moles on each side of the balanced equation (solids and liquids invisible), subtract reactants from products.

Two professional footnotes. First, strictly dimensionless thermodynamic K uses standard-state ratios (P/P° with P° = 1 bar), which absorbs the units debate into a convention — textbooks that print Kp “in atm” are using the practical form this page computes, and converting between atm-based and bar-based Kp introduces factors of 1.013^Δn that matter at the margins. Second, temperature enters twice: through the explicit RT factor AND through K itself (van’t Hoff) — converting a Kc measured at one temperature with another temperature’s RT is a category error this page can’t prevent, only frame: T here must be the temperature at which Kc was measured.

Worked Examples (Verified)

  • Ammonia (defaults): Kc 0.5, 298 K, Δn −2 → Kp = 8.36 × 10⁻⁴.
  • Mole-increasing: Kc 2.5, 500 K, Δn +1 → Kp = 102.6.
  • Δn = 0: H₂ + I₂ ⇌ 2HI → Kp = Kc at every temperature.

Kp Conversion Discipline

  • Δn counts GAS moles only — solids and pure liquids contribute zero to the subtraction.
  • T in Kelvin, and the SAME temperature the Kc belongs to.
  • R matches your pressure unit: 0.08206 (atm) or 0.08314 (bar) — the factor changes Kp by 1.3% per unit Δn.
  • Direction flips invert: converting Kp back to Kc divides by (RT)^Δn — the calculator prints the back-check so the round trip is visible.
  • Reaction written backwards or doubled? K inverts / powers accordingly BEFORE conversion — Δn belongs to the equation as written.

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Frequently Asked Questions

How do you convert Kc to Kp?

Kp = Kc × (RT)^Δn, where R = 0.08206 L·atm/mol·K, T is in Kelvin, and Δn = (moles of gaseous products) − (moles of gaseous reactants) in the balanced equation. Example (ammonia synthesis): Kc 0.5 at 298 K with Δn = −2 → Kp = 0.5 × (24.45)^(−2) = 8.36 × 10⁻⁴. Enter Kc, T and Δn above for the conversion with a back-check.

What is Δn and how do I count it?

Δn is the change in moles of GAS across the balanced equation: sum the stoichiometric coefficients of gaseous products, subtract those of gaseous reactants. N₂ + 3H₂ → 2NH₃: 2 − (1 + 3) = −2. CaCO₃(s) → CaO(s) + CO₂(g): 1 − 0 = +1 (solids ignored). H₂ + I₂ → 2HI: 2 − 2 = 0, hence Kp = Kc. Liquids and solids never enter the count.

When are Kp and Kc equal?

Exactly when Δn = 0 — the reaction neither creates nor consumes net gas moles — because (RT)⁰ = 1 at any temperature. Gas reactions with balanced gas counts (H₂ + I₂ ⇌ 2HI, CO + H₂O ⇌ CO₂ + H₂) live here. In every other case they differ by the (RT)^Δn factor, which at room temperature is ~24.5 per unit of Δn — a big lever even for Δn = ±1.

Why does temperature appear in the conversion?

Because pressure and concentration are linked by the gas law: P = cRT. At higher T the same concentration exerts proportionally more pressure, so pressure-based K drifts from concentration-based K by the (RT)^Δn factor even before K’s own temperature dependence (van’t Hoff) is considered. Both effects are real and separate: convert at the temperature of the measurement, then apply van’t Hoff reasoning if you need K at another temperature.

Is Kp unitless?

The thermodynamic K is dimensionless (activities relative to standard states: P/P° = 1 bar). The practical Kp of textbooks carries implicit pressure units from the partial pressures used. This calculator follows the practical convention with R = 0.08206 (atm-based); if your course uses bar-based Kp, multiply/divide by 1.013^Δn or equivalently use R = 0.08314 in the same formula — differences are ~1.3% per Δn unit and matter only in precise work.

Kp vs Kc — which should I use?

Whichever matches your data and your problem: gas-phase industrial and atmospheric chemistry usually report Kp (pressures are what reactors measure); solution-style treatments and most textbook tables report Kc. Equilibrium POSITION calculations must pair Q with the matching K (Qp with Kp, Qc with Kc — the Reaction Quotient page enforces the same matching). Convert once, at the start, with this formula.

Does changing pressure change Kp?

No — Kp depends only on temperature. Compressing a gas mixture changes the reaction QUOTIENT Qp (through the partial pressures), and the system shifts (Le Chatelier) until Qp again equals the unchanged Kp. The Haber-Bosch process exploits exactly this: high pressure shifts position toward ammonia (fewer gas moles) at fixed Kp. Confusing “K changes” with “position changes” is the conceptual error this distinction exists to prevent.

Is this Kp calculator free?

Yes — free, instant, no sign-up. Enter Kc, temperature in Kelvin and Δn for Kp with the conversion factor and a back-check.