Converting Kp to Kc: Equilibrium Constant Guide

Converting Kp to Kc means changing an equilibrium constant expressed in partial pressures (Kp) into one expressed in molar concentrations (Kc). You do this with the formula Kp = Kc(RT)^Δn, where Δn is the change in moles of gas. Enter your Kp, temperature, and Δn below to get Kc instantly.

Formula

The relationship between Kp and Kc is:

Kp = Kc (RT)^Δn

Where:

• Kp – equilibrium constant in terms of partial pressures (usually in atm)
• Kc – equilibrium constant in terms of molar concentrations (mol/L)
• R – ideal gas constant (0.08206 L·atm·molâ�»Â¹Â·Kâ�»Â¹ when pressure is in atm)
• T – absolute temperature in kelvin (K)
• Δn – change in moles of gas = (moles of gaseous products) − (moles of gaseous reactants)

To solve for Kc, rearrange: Kc = Kp / (RT)^Δn

Worked Example

Using the default values: Kp = 1.5 atm, T = 298 K, Δn = 1.

1. Compute RT: 0.08206 × 298 = 24.45388
2. Raise to Δn: (24.45388)^1 = 24.45388
3. Divide Kp by that: 1.5 / 24.45388 ≈ 0.06134

So Kc ≈ 0.0613 (mol/L)^1. The calculator shows 0.0613 with 4 decimal places.

Why the Conversion Matters

Chemists use Kp when dealing with gas-phase reactions measured by pressure, and Kc when working with concentrations in solution or gas concentrations in mol/L. The two constants describe the same equilibrium but in different units. Converting allows you to compare data from different sources, use equilibrium expressions in calculations that require concentrations (like acid-base or solubility problems), or simply switch between pressure-based and concentration-based frameworks.

In many textbooks, the conversion is introduced alongside the ideal gas law because the link comes from PV = nRT, which relates partial pressure to molar concentration: P = (n/V)RT = CRT. Substituting this into the Kp expression yields the Kp–Kc relationship.

How to Choose the Right Values

To use the conversion correctly, you need:

• Kp value – usually given in atm, sometimes in bar. If using bar, adjust R accordingly (R = 0.08314 L·bar·molâ�»Â¹Â·Kâ�»Â¹).
• Temperature – must be in kelvin. If given in Celsius, add 273.15.
• Δn – calculate from the balanced equation: sum of stoichiometric coefficients of gaseous products minus sum for gaseous reactants. Only count gases, not solids or liquids.

For example, for N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Δn = 2 − (1+3) = −2. For a reaction with equal moles of gas on both sides, Δn = 0, and Kp = Kc because (RT)^0 = 1.

Common Mistakes and Misconceptions

One frequent error is forgetting to convert Celsius to kelvin. Another is using the wrong value of R: if pressure is in atm, use 0.08206; if in bar, use 0.08314. Also, students often miscount Δn by including solids or liquids – only gases count. Finally, remember that Kp and Kc have different units (unless Δn = 0), so they are numerically different; don’t treat them as interchangeable without conversion.

A common misconception is that the conversion only works at standard temperature. It works at any temperature, but you must use the actual temperature of the equilibrium, not 298 K unless that’s the given temperature.

Limitations of the Formula

The Kp–Kc relationship assumes ideal gas behavior. At very high pressures or low temperatures, real gases deviate, and the conversion becomes approximate. Also, the formula applies only to gas-phase equilibria; for reactions in solution, you use Kc directly. The conversion does not change the equilibrium position – it only changes the units of the constant.

When Δn is large (e.g., ±3 or more), the numerical difference between Kp and Kc can be significant, so always check your calculation.

Frequently Asked Questions

What is the unit of Kc after conversion?

Kc has units of (mol/L)^Δn. For example, if Δn = 1, the unit is mol/L; if Δn = −1, it’s L/mol. Many textbooks omit units for simplicity, but they exist.

Can I convert Kc to Kp using the same formula?

Yes, just rearrange: Kp = Kc(RT)^Δn. The same relationship works in both directions.

What if Δn = 0?

Then (RT)^0 = 1, so Kp = Kc. This happens when the number of moles of gaseous products equals the number of moles of gaseous reactants.

Should I use R = 0.08206 or 8.314?

Use 0.08206 when pressure is in atm and volume in liters. Use 8.314 if you are working in SI units (Pa·m³), but then Kp must be in pascals. For most chemistry problems, atm is standard, so 0.08206 is the right choice.

Does the conversion depend on the stoichiometric coefficients?

Yes, because Δn comes from the balanced equation. If you double the entire equation, Δn doubles, and Kp/Kc values change accordingly (the equilibrium constant is raised to the power of the multiplier).

Why do some reactions have Kp without units?

Some sources report Kp as dimensionless by dividing each pressure by a standard pressure (like 1 atm). In that case, the conversion formula still works numerically, but the units become consistent. Always check the context of your problem.

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in kp | Kp value | atm | 1.5 | 0
in t | Temperature | K | 298 | 0
in dn | Δn (moles gas products − moles gas reactants) | - | 1 | 0
out kc | Kc value | (mol/L)^Δn | kp / ((0.08206 * t)^dn) | 4