Kp from Kc (Kp = Kc(RT)^Δn)
Worked example: Kc = 1.000, dn = +1 at 1000 K → Kp = 82.0574 — press Try an example to run it live, then adjust anything.
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Kp meets Kc →
Grade 12Grade 12 Chemistry
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Kp from Kc (Kp = Kc(RT)^Δn) explained
Gas-phase equilibria can be written with partial pressures or with molar concentrations, and the ideal gas law (P = (n/V)RT) converts between them one species at a time. Every mole of gas on the product side contributes a factor of RT and every mole on the reactant side removes one, so only the net change Δn survives: Kp = Kc. This calculator uses R = 0.08206 L·atm/(mol·K), the value that pairs partial pressures in atmospheres with concentrations in mol/L.
Δn counts gases only — solids and liquids never appear in either constant. For the Haber synthesis N₂ + 3H₂ ⇌ 2NH₃, Δn = 2 − 4 = −2, so Kp is far smaller than Kc at high temperature. Where Δn = 0, as in H₂ + I₂ ⇌ 2HI, the two constants are numerically identical and the whole conversion evaporates. A worked case: a reaction with Δn = +1 and Kc = 1.00 at 1000 K has Kp = 1.00 × (0.08206 × 1000) = 82.1 — a reminder that "the" equilibrium constant is meaningless until you say which basis you meant.
Kp from Kc (Kp = Kc(RT)^Δn) formula
- = Pressure equilibrium constant
- = Concentration equilibrium constant
- = Change in moles of gas
- = Absolute temperature (°C)
Missing one of these? Work it out first, then come back
- Pressure equilibrium constant — Gibbs Free Energy and the Equilibrium Constant, van 't Hoff Equation (K at Two Temperatures)
- Concentration equilibrium constant — Gibbs Free Energy and the Equilibrium Constant, van 't Hoff Equation (K at Two Temperatures)
- Absolute temperature — Gas Density from Molar Mass, Osmotic Pressure (Π = MRT)