Calculate Kp from Kc, or Kc from Kp, using the change in moles of gas and temperature, with results in atm, bar, kPa, mmHg and Pa.
Kc to Kp Formula
The two equilibrium constants for a gas-phase reaction are linked by the ideal gas law. Converting in either direction uses the same relationship.
Rearranged to go the other way:
The exponent comes from the balanced equation:
- Kp is the equilibrium constant written in partial pressures.
- Kc is the equilibrium constant written in molar concentrations, in mol/L.
- R is the ideal gas constant. Its value depends on the pressure unit you want Kp in: 0.0820574 L*atm/(mol*K), 0.0831446 L*bar/(mol*K), 8.31446 L*kPa/(mol*K), 62.3637 L*mmHg/(mol*K), or 8314.46 L*Pa/(mol*K).
- T is the absolute temperature in Kelvin. Celsius values must be shifted by 273.15 before use.
- delta n is the change in moles of gas across the balanced equation. Solids, liquids and aqueous species are excluded because they have no partial pressure in the equilibrium expression.
The calculator handles each piece of that formula. The solve-for selector switches between multiplying by (RT)^delta n to get Kp and dividing by it to get Kc. The delta n selector lets you type the exponent directly or build it from the gas coefficients on each side, which is safer when the equation contains solids or liquids that should be left out. The temperature field converts Celsius or Fahrenheit input to Kelvin before the exponent is applied. The pressure unit selector picks the matching value of R, so the answer comes back in the unit you actually need rather than only in atmospheres. The optional outputs report Kp in every pressure unit at once and, if you turn it on, the standard free energy change calculated from the bar value of Kp.
Gas Constants and Reference Values for the Conversion
The number you get for Kp is not unique unless you state the pressure unit. Each unit needs its own R, and the product RT is the quantity actually raised to the power delta n.
| Pressure unit for Kp | R value | RT at 298.15 K | RT at 500 K |
|---|---|---|---|
| atm | 0.0820574 L*atm/(mol*K) | 24.465 | 41.029 |
| bar | 0.0831446 L*bar/(mol*K) | 24.790 | 41.572 |
| kPa | 8.31446 L*kPa/(mol*K) | 2478.96 | 4157.23 |
| mmHg | 62.3637 L*mmHg/(mol*K) | 18593.7 | 31181.9 |
| Pa | 8314.46 L*Pa/(mol*K) | 2.479 x 10^6 | 4.157 x 10^6 |
A worked consequence: a reaction with delta n = +1 and Kc = 0.0245 mol/L at 523 K gives Kp = 1.05 atm, but the same equilibrium is Kp = 1.06 bar and Kp = 106 kPa. Textbook problems almost always assume atm, while thermodynamic tables and the expression for standard free energy assume bar.
The next table shows how delta n is read off a balanced equation, including cases where condensed phases are present.
| Balanced equation | Gas moles (products / reactants) | delta n | Kp compared with Kc (in atm, T above 12.2 K) |
|---|---|---|---|
| H2(g) + I2(g) = 2HI(g) | 2 / 2 | 0 | Equal |
| N2(g) + 3H2(g) = 2NH3(g) | 2 / 4 | -2 | Kp much smaller |
| 2SO2(g) + O2(g) = 2SO3(g) | 2 / 3 | -1 | Kp smaller |
| N2O4(g) = 2NO2(g) | 2 / 1 | +1 | Kp larger |
| CH4(g) + H2O(g) = CO(g) + 3H2(g) | 4 / 2 | +2 | Kp much larger |
| CaCO3(s) = CaO(s) + CO2(g) | 1 / 0 | +1 | Kp larger, solids ignored |
| C(s) + CO2(g) = 2CO(g) | 2 / 1 | +1 | Kp larger, solid ignored |
| NH4Cl(s) = NH3(g) + HCl(g) | 2 / 0 | +2 | Kp much larger |
The temperature threshold in the last column exists because Kp and Kc differ only through (RT)^delta n. In atm units RT passes 1 at about 12.2 K, so for any ordinary temperature a positive delta n makes Kp larger than Kc and a negative delta n makes it smaller. In kPa or Pa units RT is far above 1, so the gap is much wider for the same reaction.
Example Problems
Example 1: Kp from Kc. For N2(g) + 3H2(g) = 2NH3(g), Kc = 0.500 at 400 K. Gas products total 2 moles and gas reactants total 4 moles, so delta n = 2 – 4 = -2. Using R = 0.0820574 L*atm/(mol*K), RT = 0.0820574 * 400 = 32.823. Raising that to the power -2 gives (RT)^-2 = 9.282 x 10^-4. Then Kp = 0.500 * 9.282 x 10^-4 = 4.641 x 10^-4 atm^-2. The negative exponent pulls Kp far below Kc, which is expected for a reaction that consumes gas moles.
Example 2: Kc from Kp. For PCl5(g) = PCl3(g) + Cl2(g), Kp = 1.05 atm at 250 degrees Celsius. Convert the temperature first: 250 + 273.15 = 523.15 K. Gas products total 2 moles and gas reactants total 1 mole, so delta n = +1. RT = 0.0820574 * 523.15 = 42.928. Because delta n is +1, divide instead of multiply: Kc = 1.05 / 42.928 = 0.0245 mol/L.
FAQ
When are Kp and Kc the same number? Only when delta n equals zero. In that case (RT)^0 = 1 and the two constants are numerically identical no matter what the temperature is or which pressure unit you use. H2(g) + I2(g) = 2HI(g) is the standard example, since two moles of gas appear on each side. Any reaction that changes the gas mole count gives two different numbers.
Do solids, liquids and aqueous species count in delta n? No. Pure solids and pure liquids do not appear in the equilibrium expression at all, so they contribute nothing to delta n. Aqueous species appear in Kc but have no partial pressure, so the Kp to Kc conversion is only defined for the gaseous part of the reaction. For CaCO3(s) = CaO(s) + CO2(g), delta n is +1, not 0, because only CO2 counts. If your reaction mixes gases and aqueous species, the simple (RT)^delta n relationship does not apply.
Why does the answer change when I switch the pressure unit? Kp carries units of pressure raised to delta n. Switching from atm to kPa multiplies the numerical value by about 101.325 for every unit of delta n, so a reaction with delta n = +2 gives a Kp roughly 10,267 times larger in kPa than in atm. The physical equilibrium is unchanged. This is why exam answers should always state the unit, and why free energy calculations use bar: the thermodynamic standard state is 1 bar, so dividing Kp in bar by 1 bar makes the constant dimensionless before you apply the expression -RT ln K.