Kp to Kc Converter – Using Kp = Kc(RT)^Δn
A gas-phase equilibrium constant comes in two forms. Kc is written with concentrations in mol/L, and Kp is written with partial pressures. This Kp to Kc converter links the two with one equation. It counts Δn from the reaction you type, picks the gas constant that matches your units, and can solve for Kp, for Kc, or for the temperature at which a given pair of values agree.
Where the formula comes from
For an ideal gas, PV = nRT, so the partial pressure of each gas is P = (n/V)RT = [X]RT. Put that into the Kp expression for aA + bB ⇌ cC + dD and every pressure term becomes a concentration times RT. Collect the RT factors and you get:
Kp = Kc × (RT)^Δn, where Δn = (c + d) − (a + b)
The exponent is Δn because each mole of gas in the expression brings one factor of RT with it. Products bring them to the numerator and reactants to the denominator, so only the difference survives.
Counting Δn correctly
Δn counts gas moles only: gaseous products minus gaseous reactants, using the coefficients of the equation exactly as written. Solids and pure liquids never appear in K, so they are left out. Dissolved (aq) species are left out too, because they stay as concentrations in both forms of the constant.
N₂O₄(g) ⇌ 2NO₂(g): Δn = 2 − 1 =+1N₂(g) + 3H₂(g) ⇌ 2NH₃(g): Δn = 2 − 4 =−2CaCO₃(s) ⇌ CaO(s) + CO₂(g): Δn = 1 − 0 =+1
If you reverse a reaction, Δn changes sign. If you halve it, Δn halves too, which is how fractional values such as −½ appear. Water is the classic trap. In 2H₂(g) + O₂(g) ⇌ 2H₂O, Δn is −1 if the water is steam but −3 if it is liquid. That is why the converter asks for phases and warns when H₂O has none.
Picking the right value of R
R has to carry the same pressure unit as Kp. With Kc in mol/L, use 0.08206 L·atm/(mol·K) for atm, 0.08314 L·bar/(mol·K) for bar and 8.314 L·kPa/(mol·K) for kPa. The value 8.314 is not wrong in itself: with Kc in mol/L it gives Kp in kPa, and with Kc in mol/m³ it gives Kp in pascals.
When Kp equals Kc
If the number of gas moles is the same on both sides, Δn = 0 and (RT)^0 = 1, so Kp = Kc. H₂(g) + I₂(g) ⇌ 2HI(g) is the textbook example. For reactions like this the constant does not depend on the pressure unit or the total pressure, and temperature only matters through K itself.
atm, bar and the standard constant K°
When Δn is not zero, Kp in atm and Kp in bar are different numbers, by a factor of 1.01325^Δn. In 1982 IUPAC changed the standard pressure from 1 atm to 1 bar, so thermodynamic tables now use a 1 bar standard state. The dimensionless K° that goes into ΔG° = −RT ln K° equals Kp in bar, with any dissolved species in mol/L. You can take it straight to the Gibbs Free Energy Calculator.
Kx and Le Chatelier's principle
At a fixed temperature Kp does not change, but the mole-fraction constant does: Kx = Kp × P_total^(−Δn). If Δn is negative, raising the total pressure increases Kx, so the mixture shifts toward the side with fewer gas molecules. This is the quantitative form of Le Chatelier's rule that pressure favours fewer moles of gas.
Limits of the conversion
The relationship assumes ideal gases. At high pressure, real gases need fugacities instead of partial pressures. Kp and Kc must also describe the same reaction, written the same way, at the same temperature. Moving K to a different temperature is a job for the Van't Hoff equation. For a mixed equilibrium such as CO₂(g) ⇌ CO₂(aq), only the gas term is converted. For dissolved-gas problems, use the Henry's Law Calculator. If you only have equilibrium amounts, start with the Equilibrium Constant Calculator.