Article — Kp Equilibrium Constant Calculator (Partial Pressures)
Kp Calculator: Equilibrium Constant from Partial Pressures
Kp is the equilibrium constant of a gas-phase reaction expressed in partial pressures. For the generic reaction aA + bB ⇌ cC + dD, Kp = (P_C^c · P_D^d) / (P_A^a · P_B^b). Kp is set by temperature alone — pressure, catalysts, and inert gases never change its value, only how fast equilibrium is reached.
The calculator above handles up to two reactants and two products. Presets cover four classic equilibria: ammonia synthesis (Haber-Bosch), NO₂ dimerization, hydrogen iodide synthesis, and phosphorus pentachloride dissociation. Pressure units toggle between atm, bar, and kPa.
What is Kp?
Kp is the ratio that the partial pressures of reactants and products converge to as a reversible gas-phase reaction reaches equilibrium. Two of its properties make it powerful: it depends only on temperature for a given reaction, and it is rigorously linked to thermodynamics via ΔG° = −RT·ln(Kp).
A large Kp (≫ 1) means products dominate at equilibrium. A small Kp (≪ 1) means reactants dominate. Kp = 1 means comparable amounts of both — and ΔG° = 0, the equilibrium of a thermoneutral system.
The Kp formula
K_p = Π P_prod^ν / Π P_react^νΔG° = −RT ln(K_p)Δn = Σν_prod − Σν_react (gases only)K_p = K_c · (RT)^ΔnFor 2NO₂ ⇌ N₂O₄ at 25 °C with P(NO₂) = 0.5 atm and P(N₂O₄) = 0.8 atm:
Kp = P(N₂O₄) / P(NO₂)² = 0.8 / 0.25 = 3.2 atm⁻¹. Δn = 1 − 2 = −1, so Kp has units of inverse pressure. The reaction sits modestly toward the dimer.
Kp vs Kc — when to use which
Kc is the equivalent constant expressed in molar concentrations rather than partial pressures. The two are connected by Kp = Kc · (RT)^Δn. Three rules of thumb:
- Pure gas reactions — use Kp; partial pressures are the natural variable
- Solutions — use Kc; partial pressures do not apply
- Δn = 0 — Kp and Kc are numerically equal (e.g. H₂ + I₂ ⇌ 2HI has Δn = 0)
The Haber-Bosch process runs at 450 °C and 200 atm despite a Kp of only 0.0008 at that temperature. The trick: pressure is high enough that even a small equilibrium fraction means a useful absolute concentration of ammonia, and product is continuously condensed out — pulling the reaction forward. The combination feeds about half of all nitrogen in human food today.
How temperature shifts Kp
The van't Hoff equation links Kp at two temperatures through the standard enthalpy of reaction:
ln(Kp₂/Kp₁) = −(ΔH°/R) · (1/T₂ − 1/T₁)
For exothermic reactions (ΔH° < 0), Kp falls as temperature rises. Le Chatelier's principle puts it in plain language: heat is a "product" of an exothermic reaction, so adding heat pushes equilibrium backward toward reactants.
Concrete example. Haber synthesis: ΔH° = −92 kJ/mol. At 298 K, Kp ≈ 600 atm⁻². At 700 K, Kp ≈ 0.0008 atm⁻² — a 750,000-fold drop. Industry compromises at 450 °C because lower temperatures, while thermodynamically better, leave the reaction too slow to be useful.
Many textbooks drop the units of Kp by quietly dividing every pressure by a 1 bar reference state. That works if you stay consistent. Comparing tabulated Kp values from different sources requires checking the reference state. ΔG° = −RT·ln(Kp) needs a dimensionless Kp built from activities, not raw pressures.
Kp and the reaction quotient Q
Q has the same algebraic form as Kp but uses non-equilibrium pressures. Comparing Q to Kp tells you which way the reaction will move:
- Q < Kp — too few products; net forward reaction (→)
- Q = Kp — at equilibrium; no net change
- Q > Kp — too many products; net reverse reaction (←)
For the same Haber system, starting with 1 atm N₂, 3 atm H₂, and 0.01 atm NH₃: Q = (0.01)² / (1 · 27) = 3.7 × 10⁻⁶. With Kp = 0.0008 at 700 K, Q ≪ Kp and the forward reaction proceeds. That gradient is what the catalyst exploits.
Industrial Kp examples (Haber, contact process)
Industry rarely runs at thermodynamic optimum. The Haber process trades a worse Kp for usable kinetics; steam reforming runs hot and accepts the energy cost. Each design balances Kp, rate, catalyst cost, and product separation.
Common Kp mistakes
The frequent errors:
- Including solids and pure liquids — they have unit activity; their "partial pressure" is not in the Kp expression
- Wrong sign of Δn — Δn = products − reactants, not the other way
- Mixing units — all pressures in atm, or all in bar; never mix
- Confusing Kp with rate constant — Kp tells you the equilibrium position, not the speed of approach
- Using non-equilibrium pressures — you computed Q, not Kp; only equilibrium pressures give the true constant
- Ignoring temperature — quoting "Kp = 4" without temperature is meaningless
If a problem gives equilibrium concentrations in mol/L instead of partial pressures, you can either compute Kc directly or convert to partial pressures via P = (mol/L)·RT — handy when only one form is tabulated for your reaction. With R = 0.0821 L·atm/(mol·K) and T in kelvin, a 1 mol/L concentration at 298 K corresponds to a 24.5 atm partial pressure.
The Haber-Bosch synthesis of ammonia consumes about 1 percent of all global energy and supplies the nitrogen fertilizer that feeds nearly half of humanity. Every percentage point of conversion efficiency is worth roughly a billion dollars per year. A century of catalyst research and reactor engineering has chased a thermodynamic constraint that started as a Kp value in 1909 — Fritz Haber's original measurement gave Kp ≈ 0.013 at 723 K and 200 atm, and that single number set the modern fertilizer industry in motion.
The lesson generalizes. Sulfuric acid production (contact process), nitric acid production (Ostwald), methanol synthesis, and gas-to-liquid Fischer-Tropsch all start from a Kp-driven feasibility analysis. The constant tells you what is thermodynamically possible; catalysis tells you what is practically achievable. Designs that ignore Kp end up either failing physically or wasting energy chasing impossible conversions.