For reactions between gases, the equilibrium constant can be written with partial pressures instead of concentrations. Kp uses the equilibrium partial pressure of each gas raised to its coefficient, while Kc uses molar concentrations. Only gases appear in Kp; solids and liquids are left out, just as in Kc.
The two constants are linked through the ideal gas law. Because p = (n/V)·RT = c·RT for each gas, Kp = Kc·(RT)^Δn, where Δn is the change in the number of moles of gas from reactants to products. If the number of gas molecules does not change (Δn = 0), as in H₂ + I₂ ⇌ 2 HI, Kp and Kc are equal. When Δn ≠ 0 the value of R must match the pressure unit: 0.08206 L·atm/(mol·K) for atmospheres, 0.08314 L·bar/(mol·K) for bar, 8.314 L·kPa/(mol·K) for kilopascals or 62.36 L·Torr/(mol·K) for Torr.
Use this calculator to convert Kc to Kp, Kp to Kc, or to compute both straight from measured partial pressures – for example in the Haber process, where Δn = −2, or the N₂O₄ ⇌ 2 NO₂ equilibrium, where Δn = +1.
How to use the Kp Calculator
- Calculate
- Kc (concentrations in mol/L)
- Kp
- Δn – change in moles of gas (products − reactants)
- Temperature
- Pressure unit of Kp
- Number of gaseous reactants
- Number of gaseous products
- Reactant 1 – coefficient
- Reactant 1 – partial pressure
- Reactant 2 – coefficient
- Reactant 2 – partial pressure
- Product 1 – coefficient
- Product 1 – partial pressure
- Product 2 – coefficient
- Product 2 – partial pressure
Results update as you type, every field has its own unit menu, and everything is calculated in your browser.
How it works
Pick the conversion and the pressure unit of Kp. The calculator selects the matching gas constant, raises RT to the power Δn and multiplies or divides. In the partial-pressure mode it converts each pressure to the chosen unit, builds Kp from the coefficients and derives Δn from them (all species are assumed to be gases).
Worked example
Kc to Kp: for N₂ + 3 H₂ ⇌ 2 NH₃ with Kc = 0.060 at 500 K, Δn = 2 − 4 = −2 and RT = 0.082057 × 500 = 41.03 L·atm/mol. Kp = 0.060 × 41.03⁻² = 0.060 ÷ 1683.4 = 3.56 × 10⁻⁵ (atm); in bar it is 3.47 × 10⁻⁵.
From partial pressures: for N₂O₄ ⇌ 2 NO₂ at 298.15 K with p(N₂O₄) = 0.20 atm and p(NO₂) = 0.40 atm, Kp = 0.40² ÷ 0.20 = 0.80, Δn = +1 and Kc = 0.80 ÷ (0.082057 × 298.15) = 0.80 ÷ 24.47 ≈ 0.0327.
Gas constant in different units
| Pressure unit | R | Unit of R |
|---|---|---|
| atm | 0.082057 | L·atm/(mol·K) |
| bar | 0.083145 | L·bar/(mol·K) |
| kPa | 8.3145 | L·kPa/(mol·K) |
| Torr (mmHg) | 62.364 | L·Torr/(mol·K) |
| Pa (SI) | 8.3145 | J/(mol·K) = m³·Pa/(mol·K) |
Tips
- Check the equation is balanced before using coefficients.
- Find the limiting reagent before calculating a yield.
- Keep temperatures in kelvin for rate and equilibrium equations.
- Everything is calculated in your browser.
Common mistakes to avoid
- Using the excess reagent to calculate theoretical yield.
- Including pure solids and liquids in equilibrium expressions.
- Entering temperatures in °C into the Arrhenius equation without converting.
Inputs explained
| Field | What to enter |
|---|---|
| Calculate | Choose one: Kp from Kc, Kc from Kp, Kp and Kc from equilibrium partial pressures |
| Kc (concentrations in mol/L) | A number |
| Kp | A number |
| Δn – change in moles of gas (products − reactants) | A number |
| Temperature | Units: K, °C, °F |
| Pressure unit of Kp | Choose one: atm – R = 0.082057 L·atm/(mol·K), bar – R = 0.083145 L·bar/(mol·K), kPa – R = 8.3145 L·kPa/(mol·K), Torr (mmHg) – R = 62.364 L·Torr/(mol·K) |
| Number of gaseous reactants | Choose one: 1, 2 |
| Number of gaseous products | Choose one: 1, 2 |
| Reactant 1 – coefficient | A number |
| Reactant 1 – partial pressure | Units: atm, kPa, Pa, bar, mmHg (Torr), psi |
| Reactant 2 – coefficient | A number |
| Reactant 2 – partial pressure | Units: atm, kPa, Pa, bar, mmHg (Torr), psi |
| Product 1 – coefficient | A number |
| Product 1 – partial pressure | Units: atm, kPa, Pa, bar, mmHg (Torr), psi |
| Product 2 – coefficient | A number |
| Product 2 – partial pressure | Units: atm, kPa, Pa, bar, mmHg (Torr), psi |
What each result means
- Kc (concentrations in mol/L) = [Kp from pressures]÷([Gas constant]×[Temperature])^[Δn from coefficients]
- Δn (moles of gas) = [Δn used]
- Conversion factor (RT)^Δn = ([Gas constant]×[Temperature])^[Δn used]
- Gas constant R used (L·pressure unit/(mol·K)) = [Gas constant]
- R·T = [Gas constant]×[Temperature]
Formula
Kp = Kc·(R·T)^Δn · Δn = moles of gaseous products − moles of gaseous reactants · Kp = Π(p_products)^ν ÷ Π(p_reactants)^νWorked example
Inputs — Calculate: Kp from Kc; Kc (concentrations in mol/L): 0.06; Δn – change in moles of gas (products − reactants): -2; Temperature: 500 K; Pressure unit of Kp: atm – R = 0.082057 L·atm/(mol·K).
Results — Kp = 0.000036; Δn (moles of gas) = -2; Conversion factor (RT)^Δn = 0.000594; Gas constant R used (L·pressure unit/(mol·K)) = 0.082057; R·T = 41.0287.
Key terms
- Kp
- Equilibrium constant written with partial pressures of gases.
- Kc
- Equilibrium constant written with molar concentrations.
- Δn
- Change in the number of moles of gas from reactants to products in the balanced equation.
- Partial pressure
- The pressure one gas in a mixture would exert if it filled the container alone.
Frequently asked questions
What is the formula to convert Kc to Kp?
Kp = Kc·(RT)^Δn, with T in kelvin, R matched to the pressure unit and Δn = moles of gaseous products − moles of gaseous reactants.
When are Kp and Kc equal?
When Δn = 0 – the same number of gas molecules on both sides, as in H₂ + I₂ ⇌ 2 HI or N₂ + O₂ ⇌ 2 NO.
Which value of R should I use?
0.082057 L·atm/(mol·K) for Kp in atm, 0.083145 L·bar/(mol·K) for bar, 8.3145 L·kPa/(mol·K) for kPa and 62.364 L·Torr/(mol·K) for Torr. Using 8.314 J/(mol·K) with pressures in atm gives wrong answers.
How do I find Δn?
Count only gases. For N₂(g) + 3 H₂(g) ⇌ 2 NH₃(g), Δn = 2 − 4 = −2; for CaCO₃(s) ⇌ CaO(s) + CO₂(g), Δn = 1 − 0 = +1.
What is Kp for the Haber process if Kc = 0.06 at 500 K?
Kp = 0.06 × (0.082057 × 500)⁻² = 0.06 ÷ 1683.4 ≈ 3.56 × 10⁻⁵ (pressures in atm).
Are solids and liquids included in Kp?
No – only gases. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp is simply the equilibrium pressure of CO₂.
Does the pressure unit change Kp?
Yes when Δn ≠ 0. Kp in bar and Kp in atm differ by a factor 1.01325^Δn; with Δn = 0 the value is the same in any unit.
What formula does the Kp Calculator use?
Kp = Kc·(R·T)^Δn · Δn = moles of gaseous products − moles of gaseous reactants · Kp = Π(p_products)^ν ÷ Π(p_reactants)^ν.