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Kp Calculator — Convert Kc to Kp and Kp to Kc | LazyTools
Math & Science

Kp Calculator (Kc ⇌ Kp Conversion)

Convert between Kc (concentration-based equilibrium constant) and Kp (pressure-based) using Kp = Kc × (RT)^Δn. Furthermore, Δn is the change in moles of gaseous species (gas product moles minus gas reactant moles), and R = 0.08206 L·atm/mol·K.

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How to use the Kp Calculator (Kc ⇌ Kp Conversion)

1
Select direction

Choose Kc → Kp or Kp → Kc. Furthermore, both directions use the same equation Kp = Kc × (RT)^Δn, just rearranged.

2
Enter Kc or Kp

Type the known equilibrium constant. Furthermore, Kc uses mol/L concentrations; Kp uses atm pressures.

3
Enter Δn

Δn = moles of gaseous products − moles of gaseous reactants in the balanced equation. Furthermore, pure solids, liquids, and solvents are excluded.

4
Enter temperature

Required because RT varies with T. Furthermore, use Kelvin or Celsius — the calculator converts.

5
Read the result

Kp or Kc appears alongside (RT)^Δn conversion factor. Moreover, when Δn = 0, Kp = Kc exactly.

Variants, options and when to use each

ReactionΔnKp vs Kc
N₂(g)+3H₂(g)⇌2NH₃(g)-2Kp < Kc (Δn negative)
N₂O₄(g)⇌2NO₂(g)+1Kp > Kc
H₂(g)+I₂(g)⇌2HI(g)0Kp = Kc
CO(g)+3H₂(g)⇌CH₄(g)+H₂O(g)-2Kp < Kc

The formula explained

Kp = Kc × (RT)^Δn | Kc = Kp / (RT)^Δn (R = 0.08206 L·atm/mol·K)
Kp = pressure equilibrium constant (dimensionless, using P/P° where P° = 1 atm)
Kc = concentration equilibrium constant (dimensionless, using c/c° where c° = 1 mol/L)
R = 0.08206 L·atm/mol·K (gas constant in pressure-volume units)
T = temperature (Kelvin)
Δn = (mol gaseous products) − (mol gaseous reactants)

The relationship Kp = Kc(RT)^Δn comes from PV = nRT → P = (n/V)RT = [c]RT for ideal gases. Furthermore, each gaseous species contributes a factor of RT when converting from concentration to pressure. Δn net factors of RT emerge — one for each net mole of gas formed. Moreover, when Δn = 0, no net RT factors appear and Kp = Kc exactly.

Worked example — N₂O₄ ⇌ 2NO₂ at 25°C

StepCalculationResult
Kc at 25°Cgiven4.61 × 10⁻³
Δn = 2 − 1mol NO₂ products − mol N₂O₄+1
RT = 0.08206 × 29824.45
Kp = Kc × (RT)^14.61×10⁻³ × 24.450.1127
Kp = 0.1127 for N₂O₄ ⇌ 2NO₂ at 25°C — significantly larger than Kc = 4.61×10⁻³. Furthermore, Δn = +1 gives one factor of RT = 24.45 that multiplies Kc. Moreover, this illustrates why Kp > Kc when Δn > 0 — more gas moles means higher pressure at equilibrium relative to concentration.

What are Kp and Kc in equilibrium chemistry?

Kc is the equilibrium constant expressed in terms of molar concentrations (mol/L). Kp is the equilibrium constant expressed in terms of partial pressures (atm). Furthermore, both describe the same equilibrium — they differ only in the units used to express species amounts. For reactions involving no gases or equal gas moles on both sides (Δn = 0), Kp = Kc.

The conversion factor (RT)^Δn comes from the ideal gas law PV = nRT. Moreover, for a gas, concentration c = n/V = P/(RT), so P = cRT. Each gaseous species in the equilibrium expression contributes a factor of RT when converting between concentration and pressure. Moreover, Δn net factors of RT appear in Kp/Kc.

Thermodynamic equilibrium constants are formally dimensionless — concentrations are divided by a standard state (c° = 1 mol/L) and pressures by a standard state (P° = 1 atm). Additionally, this is why Kp and Kc have the same numerical value when using these dimensionless conventions — the RT conversion is built into the units.

Who uses this calculator?

Physical chemists convert between Kp and Kc when equilibrium data is available in one form but needed in another. Furthermore, gas-phase equilibria are more naturally described by Kp; aqueous equilibria by Kc. Chemical engineers designing gas-phase reactors (Haber process, SO₃ oxidation, steam reforming) use Kp with partial pressures from process conditions. Moreover, thermodynamic tables typically list Kp values for gas reactions.

Historical context and related concepts

The equilibrium constant was introduced by Guldberg and Waage (1864) in concentration form. Furthermore, the pressure form Kp and the relationship Kp = Kc(RT)^Δn were developed as kinetic theory and the ideal gas law were connected to thermodynamic equilibrium in the late 19th century. Van't Hoff's equation (1884) related K to temperature, establishing the complete thermodynamic framework. Moreover, modern thermodynamics defines K in terms of dimensionless activities — unifying Kp and Kc.

Why Kp is essential for industrial gas-phase equilibrium design

Industrial gas reactions (ammonia synthesis, methanol production, Fischer-Tropsch synthesis) are designed around Kp values at specific temperatures and pressures. Furthermore, increasing pressure shifts equilibria with Δn < 0 toward products — the Haber process runs at 150–300 atm precisely because Δn = −2 for NH₃ synthesis. Moreover, Kp calculations guide the selection of operating temperature and pressure to achieve target conversion.

Kp in atmospheric chemistry and ozone depletion

Atmospheric equilibria for ozone (O₃ ⇌ O₂ + O, Δn = +1), NO₂ (2NO₂ ⇌ N₂O₄, Δn = −1), and hydroxyl radical reactions are all described by Kp values at stratospheric temperatures. Furthermore, the ozone hole results from catalytic Cl reactions with Kp-governed kinetics — temperature and pressure determine both Kp and rate constants. Moreover, atmospheric models require precise Kp values to simulate photochemical smog and climate chemistry.

Frequently asked questions

Δn = (sum of stoichiometric coefficients of gaseous products) − (sum of coefficients of gaseous reactants). Furthermore, pure solids, pure liquids, and solvent molecules (treated as constant concentration) are excluded from both sides. Example: N₂(g) + 3H₂(g) → 2NH₃(g): Δn = 2 − (1+3) = −2.
When Δn = 0 — equal moles of gas on both sides. Furthermore, Example: H₂(g) + I₂(g) ⇌ 2HI(g): Δn = 2 − (1+1) = 0, so Kp = Kc. Moreover, reactions involving only solids, liquids, or aqueous species (no gas) effectively have Δn = 0 for Kp/Kc comparison purposes.
R = 0.08206 L·atm/mol·K gives Kp in atm units. Furthermore, if Kp is defined using pressures in Pa or bar, use R = 8.314 J/mol·K (Pa) or R = 0.08314 L·bar/mol·K. Moreover, the calculated Kp depends on the pressure units chosen for Kp — always state the units convention.
Kp changes with temperature according to van't Hoff equation: d(lnKp)/dT = ΔH°/(RT²). Furthermore, exothermic reactions (ΔH° < 0) have decreasing Kp with increasing T; endothermic reactions (ΔH° > 0) have increasing Kp. Moreover, this determines optimal temperature for industrial processes — low T for exothermic (more product), high T for endothermic, balanced against reaction rate.
Thermodynamic K is dimensionless and uses activities (a = c/c° for concentrations, a = P/P° for pressures). Furthermore, it equals Kc when c° = 1 mol/L and Kp when P° = 1 atm — making both conventional K values numerically equal to the thermodynamic K in appropriate units. Moreover, ΔG° = −RTlnK always uses the dimensionless thermodynamic K.

Related tools

Equilibrium Constant Calculator

Calculate K from ΔG°. Furthermore, Kp and Kc are interconverted using Kp = Kc(RT)^Δn.

Reaction Quotient Calculator

Calculate Q and compare to K. Moreover, Qp and Qc follow the same (RT)^Δn relationship as Kp and Kc.

Ideal Gas Law Calculator

PV = nRT is the basis of Kp = Kc(RT)^Δn. Furthermore, the conversion factor comes directly from ideal gas law.

Gibbs Free Energy Calculator

ΔG° = −RTlnK. Moreover, both Kp and Kc give the same ΔG° when used consistently with their standard states.

Combined Gas Law Calculator

Analyse gas-phase conditions. Additionally, partial pressures used in Kp calculations follow combined gas law.

Significant Figures Calculator

Round K values to appropriate precision. Furthermore, equilibrium constants are typically precise to 2–3 significant figures.

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