Kp to Kc Calculator
Convert an equilibrium constant between its pressure-based form (Kp) and concentration-based form (Kc). Enter the known constant, the temperature, and the change in moles of gas to get the result instantly.
Convert Between Pressure- and Concentration-Based Equilibrium Constants
Kp to Kc Formula
Where:
- \(K_p\) — equilibrium constant based on partial pressures (atm)
- \(K_c\) — equilibrium constant based on molar concentrations (mol/L)
- \(R\) — gas constant, 0.082057 L·atm/(mol·K)
- \(T\) — absolute temperature (K)
- \(\Delta n\) — change in moles of gas: \(\Delta n = \) (moles of gaseous products) \(-\) (moles of gaseous reactants)
The two constants describe the same equilibrium but are expressed in different units; the factor \((RT)^{\Delta n}\) converts between them.
Kp to Kc – Calculation Example
Example: for the reaction \(2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g)\), the equilibrium constant is \(K_c = 4.0\) at 500 K. Find Kp.
Count only gaseous species: \(\Delta n = 2 - (2 + 1) = -1\)
\[ K_p = K_c \cdot (RT)^{\Delta n} = 4.0 \times (0.082057 \times 500)^{-1} = 4.0 \times (41.0285)^{-1} \]
\[ K_p = 4.0 \times 0.024373 = 0.097494 \]
Result: \(K_p \approx 0.0975\) at 500 K.
How to Use the Kp to Kc Calculator
- Open the What to Calculate dropdown and choose whether you want Kp (from a known Kc) or Kc (from a known Kp).
- Enter your value in Known Equilibrium Constant — the Kc if you are finding Kp, or the Kp if you are finding Kc.
- Type the temperature in Temperature and select the unit (K or °C) from the dropdown next to it. The calculator converts °C to kelvin automatically.
- Enter Change in Moles of Gas: subtract the total moles of gaseous reactants from the total moles of gaseous products. This value can be negative, zero, or positive.
- Read the results: Kp, Kc, the conversion factor \((RT)^{\Delta n}\), and a step-by-step calculation trail showing exactly how the answer was obtained.
What Are Kp and Kc?
Chemical equilibria involving gases can be described in two equivalent ways. \(K_c\) is built from the equilibrium molar concentrations of the species (mol/L), while \(K_p\) is built from their partial pressures (atm). Textbooks and exam problems use both, and it is common to be given one form and asked for the other.
The link between them comes from the ideal gas law, \(PV = nRT\), which relates a gas's partial pressure to its concentration: \(P = (n/V)RT\). When you substitute this into the equilibrium expression, every gaseous species contributes a factor of \(RT\), and these factors collect into \((RT)^{\Delta n}\), where \(\Delta n\) is the change in the number of moles of gas between products and reactants.
The sign of \(\Delta n\) controls the direction of the conversion. If the reaction produces more gas molecules than it consumes (\(\Delta n > 0\)), then \(K_p > K_c\) at typical temperatures. If it consumes gas (\(\Delta n < 0\)), then \(K_p < K_c\). If the number of gas molecules is unchanged (\(\Delta n = 0\)), the two constants are numerically equal — for example, for \(H_2(g) + I_2(g) \rightleftharpoons 2HI(g)\).
Keep in mind that neither constant is truly dimensionless in this elementary treatment; they are conventionally reported as plain numbers. The conversion factor \((RT)^{\Delta n}\) carries the units, which is why the numerical value of the constant changes between the two forms.
Assumptions, Limitations, and Tips
- Count only gaseous species. Pure solids and liquids do not appear in equilibrium expressions, so they must be excluded when computing \(\Delta n\). For \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\), \(\Delta n = 1\), not 2 − 1 = 1 from all species counted.
- The gas constant must match the units. This calculator uses \(R = 0.082057\) L·atm/(mol·K), which is the correct choice when pressures are in atm and concentrations in mol/L. If a problem quotes pressures in bar or kPa, convert them to atm first or use the corresponding value of R.
- Temperature must be absolute. The formula requires kelvin. If you enter °C, the calculator adds 273.15 for you, but double-check that you selected the right unit.
- Δn must be a whole number. Count the stoichiometric coefficients of gaseous products and reactants from the balanced equation; do not use the actual amounts present at equilibrium.
- Ideal behavior is assumed. At very high pressures or low temperatures, real gases deviate from ideality and the simple \((RT)^{\Delta n}\) relationship becomes approximate.
- Double-check your balanced equation first — a wrong \(\Delta n\) is the most common source of error in these conversions.
Quick Reference: Δn for Common Reaction Types
| Reaction (gases only counted) | \(\Delta n\) | Relation at 500 K |
|---|---|---|
| \(N_2 + 3H_2 \rightleftharpoons 2NH_3\) | \(-2\) | \(K_p = K_c / (RT)^2 \approx K_c / 1683\) |
| \(2SO_2 + O_2 \rightleftharpoons 2SO_3\) | \(-1\) | \(K_p \approx K_c / 41.03\) |
| \(H_2 + I_2 \rightleftharpoons 2HI\) | \(0\) | \(K_p = K_c\) |
| \(N_2O_4 \rightleftharpoons 2NO_2\) | \(+1\) | \(K_p \approx K_c \times 41.03\) |
| \(PCl_5 \rightleftharpoons PCl_3 + Cl_2\) | \(+1\) | \(K_p \approx K_c \times 41.03\) |
To explore the equilibrium concept further, try our Equilibrium Constant calculator, or connect \(K\) to thermodynamics with the Gibbs Free Energy calculator. For equilibria at electrodes, the Nernst Equation calculator is another useful tool.
Frequently Asked Questions
How do I convert Kp to Kc by hand without a calculator?
Write down the balanced equation and find \(\Delta n\) from gaseous species only. Convert the temperature to kelvin, then compute \(RT\) using \(R = 0.082057\) L·atm/(mol·K). Raise \(RT\) to the power \(\Delta n\). To go from Kc to Kp, multiply by \((RT)^{\Delta n}\); to go from Kp to Kc, divide by it. For example, with \(T = 298\) K and \(\Delta n = 1\), the factor is about 24.45.
What is a normal or good value for an equilibrium constant?
There is no universal "good" value — it depends on the reaction. As a rule of thumb, \(K > 10^3\) means products dominate at equilibrium, \(K < 10^{-3}\) means reactants dominate, and values near 1 indicate a genuine mixture of both. A very large or very small K simply tells you which side the equilibrium favors; it says nothing about how fast the reaction reaches equilibrium.
Can I use bar or kPa instead of atm with this formula?
Not directly. The relation \(K_p = K_c(RT)^{\Delta n}\) with \(R = 0.082057\) assumes pressures in atm and concentrations in mol/L. If your pressures are in bar, use \(R = 0.08314\) L·bar/(mol·K) instead; for kPa use \(R = 8.314\) L·kPa/(mol·K). Alternatively, convert all pressures to atm first (1 bar ≈ 0.98692 atm) and keep the standard formula unchanged.
Why does the Kp/Kc ratio depend on temperature?
The conversion factor \((RT)^{\Delta n}\) grows with temperature whenever \(\Delta n \neq 0\). At higher T, the gap between Kp and Kc widens for reactions that change the number of gas molecules. Note that this is separate from the fact that the equilibrium constant itself changes with temperature — both effects happen at once, so always use the temperature at which the given constant was measured.
Do solids and liquids affect Δn or the equilibrium expression?
No. Pure solids and pure liquids have constant composition, so they are omitted from equilibrium expressions entirely. For \(CaCO_3(s) \rightleftharpoons CaO(s) + CO_2(g)\), only the gaseous \(CO_2\) counts, giving \(\Delta n = 1\). Similarly, in esterification reactions the water may be excluded if it is the solvent. Counting non-gaseous species in \(\Delta n\) is one of the most frequent errors students make.
What are the most common mistakes when converting Kp and Kc?
The top errors are: using Celsius instead of kelvin; forgetting the sign of \(\Delta n\) (products minus reactants, never the reverse); counting solids or liquids; using an unbalanced equation; and mixing up the direction of conversion — Kp equals Kc times \((RT)^{\Delta n}\), not divided by it. Also check whether the problem uses 0.0821 or 8.314 for R; a mismatch with the pressure units shifts the answer significantly.
How is Kp related to Gibbs free energy?
The thermodynamic (dimensionless) equilibrium constant K connects to free energy via \(\Delta G^\circ = -RT_{thermo}\ln K\), where \(R = 8.314\) J/(mol·K). If Kp is expressed with partial pressures divided by the standard pressure (1 atm), it is essentially this thermodynamic K, so you can plug it directly into the equation. A negative \(\Delta G^\circ\) corresponds to \(K > 1\), meaning products are favored under standard conditions.