Kp Calculator: Equilibrium Constant from Partial Pressures
Kp is the equilibrium constant expressed in terms of the partial pressures of gases in a reaction mixture. Use this calculator in two ways: enter Kc, temperature, and the change in moles of gas to convert to Kp, or enter the partial pressures of each reactant and product directly to compute Kp from first principles. The result updates instantly as you type.
Formula
Worked example
For N2(g) + 3H2(g) = 2NH3(g): Kc = 0.0227 at 298 K. delta-n = 2 - (1+3) = -2. R = 0.08206 L atm/(K mol). RT = 0.08206 * 298 = 24.45. (RT)^(-2) = 1/(24.45^2) = 0.001673. Kp = 0.0227 * 0.001673 = 3.80e-5.
What is Kp and how does it differ from Kc?
Every chemical equilibrium has an equilibrium constant that describes the ratio of products to reactants when the forward and reverse reaction rates become equal. Kc expresses this ratio using molar concentrations (mol/L), whereas Kp uses the partial pressures of each gaseous species. For reactions that involve only dissolved species or pure solids and liquids, Kc is the natural choice. For gas-phase reactions, Kp is often more convenient because pressures are easier to measure directly in a sealed vessel. Both constants carry exactly the same thermodynamic information; they differ only in units and by the factor (RT)^(delta-n). When delta-n = 0, meaning the same number of gas moles appear on each side of the equation, Kp equals Kc exactly.
The Kp formula and how to use it
For a general gas-phase reaction aA + bB = cC + dD (where a, b, c, d are stoichiometric coefficients), Kp is defined as Kp = (PC^c * PD^d) / (PA^a * PB^b), where each P is the equilibrium partial pressure of that species. Only gaseous species appear in the expression; pure solids and liquids are excluded because their "concentrations" are constant and are absorbed into the equilibrium constant. The relationship between Kp and Kc is Kp = Kc * (RT)^(delta-n), where R is the ideal gas constant in appropriate units, T is the absolute temperature in Kelvin, and delta-n is the change in the total number of moles of gas across the reaction. The value of R depends on the pressure unit you choose: 0.08206 L atm/(K mol) for atmospheres, 8.314 L kPa/(K mol) for kilopascals, or 0.08314 L bar/(K mol) for bar.
Interpreting the magnitude of Kp
A Kp much greater than 1 means the reaction lies strongly toward products at equilibrium: the partial pressures of products greatly exceed those of reactants. A Kp much less than 1 means the reverse is true and reactants dominate at equilibrium. A Kp close to 1 indicates that products and reactants are present in comparable amounts. In many industrial processes the goal is a large Kp, and chemists adjust temperature and pressure to push the value in the desired direction. For exothermic reactions, lowering the temperature increases Kp; for endothermic reactions, raising the temperature increases Kp. This is a direct consequence of the van't Hoff equation, which links the temperature dependence of Kp to the enthalpy change of the reaction.
Worked example: ammonia synthesis
The Haber process for making ammonia is N2(g) + 3H2(g) = 2NH3(g). At 298 K, Kc = 2.27 * 10^(-2). The change in gas moles is delta-n = 2 - (1 + 3) = -2. Using R = 0.08206 L atm/(K mol) and T = 298 K: RT = 24.45 L atm/mol. (RT)^(-2) = 1 / (24.45^2) = 0.001673. Therefore Kp = 0.0227 * 0.001673 = 3.80 * 10^(-5). This very small Kp says that at room temperature the partial pressure of ammonia at equilibrium is tiny compared to the reactant pressures. Industrially the reaction is run at high pressure and moderate temperature (around 400-500 degrees C) to get a commercially useful yield.
Gas constant R values by pressure unit
| Pressure unit | R value | Units of R |
|---|---|---|
| atm | 0.08205746 | L*atm/(K*mol) |
| kPa | 8.3144621 | L*kPa/(K*mol) |
| bar | 0.08314462 | L*bar/(K*mol) |
| Torr | 62.36367 | L*Torr/(K*mol) |
| mmHg | 62.36367 | L*mmHg/(K*mol) |
Use the R value that matches your chosen pressure unit when applying Kp = Kc * (RT)^(delta-n).
Frequently asked questions
What is Kp in chemistry?
Kp is the equilibrium constant for a gas-phase reaction expressed in terms of the partial pressures of each gaseous reactant and product. It is defined as the product of the equilibrium partial pressures of gaseous products (each raised to its stoichiometric coefficient) divided by the equivalent product for reactants. Like all equilibrium constants, Kp is dimensionless (pressures are divided by the standard pressure of 1 atm or 1 bar) and depends only on temperature.
How do you convert Kc to Kp?
Use the equation Kp = Kc * (RT)^(delta-n), where R is the ideal gas constant, T is the absolute temperature in Kelvin, and delta-n is the change in the number of moles of gas in the reaction (moles of gaseous products minus moles of gaseous reactants). Choose the value of R that matches your pressure unit: 0.08206 L atm/(K mol) for atm, 8.314 L kPa/(K mol) for kPa, or 0.08314 L bar/(K mol) for bar.
What does it mean when Kp is greater than 1?
A Kp greater than 1 means that at equilibrium the partial pressures of products are larger than those of reactants (when each is raised to its stoichiometric coefficient). The reaction favors product formation. The larger the value, the further the equilibrium lies toward complete conversion. A Kp much less than 1 means the opposite: reactants dominate at equilibrium.
Does Kp change with pressure?
No. Kp depends only on temperature, not on the total pressure of the system. Changing total pressure shifts the equilibrium position (which way the reaction goes to re-establish equilibrium), but it does not change the numerical value of Kp. The temperature dependence of Kp is described by the van't Hoff equation: for exothermic reactions Kp decreases as temperature rises; for endothermic reactions it increases.
Which species are excluded from the Kp expression?
Pure solids and pure liquids are excluded from the Kp expression because their partial pressures (or more precisely, their activities) remain essentially constant throughout the reaction and are incorporated into the equilibrium constant itself. Only gaseous species appear. For example, in CaCO3(s) = CaO(s) + CO2(g), Kp equals simply the partial pressure of CO2 at equilibrium, because the two solids have fixed activities of 1.
When does Kp equal Kc?
Kp equals Kc whenever delta-n = 0, that is, when the number of moles of gaseous products equals the number of moles of gaseous reactants. In that case (RT)^0 = 1 and the two constants are identical. A common example is H2(g) + I2(g) = 2HI(g), where delta-n = 2 - 2 = 0.