Kc to Kp Equation: Conversion & Examples

The Kc to Kp equation lets you convert an equilibrium constant expressed in terms of concentration (Kc) into one expressed in terms of partial pressure (Kp). You enter your Kc value, the temperature, and the change in gas moles (Δn), and the calculator returns the corresponding Kp using the formula Kp = Kc(RT)^Δn.

Formula

The relationship between Kc and Kp is:

Kp = Kc (RT)^Δn

• Kp – equilibrium constant in terms of partial pressures (units depend on Δn)
• Kc – equilibrium constant in terms of molar concentrations (mol/L)
• R – ideal gas constant. Use 0.08206 L·atm/(mol·K) when pressures are in atm.
• T – absolute temperature in kelvin (K)
• Δn – change in moles of gas = (moles of gaseous products) − (moles of gaseous reactants) from the balanced equation

The exponent Δn is crucial: if the number of gas molecules doesn’t change during the reaction, then Δn = 0 and Kp equals Kc (because anything raised to the power 0 is 1).

Worked Example

Using the calculator defaults: Kc = 0.5, T = 298 K, Δn = 2.

Step 1: Identify R = 0.08206 L·atm/(mol·K).

Step 2: Compute RT = 0.08206 × 298 = 24.45388.

Step 3: Raise to Δn: (24.45388)² = 597.99.

Step 4: Multiply by Kc: 0.5 × 597.99 = 299.0.

So Kp ≈ 299 (with units atm²). The calculator shows 299.0.

When to Use Kc vs Kp

Use Kc when you are working with concentrations (mol/L) – common in solution chemistry and when all species are in the same phase. Use Kp when you are dealing with gases and you have partial pressures (usually in atm or bar). Many textbooks present gas‑phase reactions with Kp because pressures are easier to measure directly.

If a reaction involves both gases and solids or liquids, remember that pure solids and pure liquids do not appear in the equilibrium expression – their “concentrations� are constant. Only gaseous species contribute to Δn.

How to Choose the Correct R Value

The value of R must match the pressure unit in your Kp. The most common pairing is:

• atm: R = 0.08206 L·atm·molâ�»Â¹Â·Kâ�»Â¹
• bar: R = 0.08314 L·bar·molâ�»Â¹Â·Kâ�»Â¹
• Pa: R = 8.314 J·molâ�»Â¹Â·Kâ�»Â¹ (but then pressures are in Pa, which is rarely used for Kp)

If you mix units – for example using R in atm but your pressures are in bar – you will get a wrong Kp. Always check that the pressure unit in R matches the pressure unit you intend to use for Kp.

Common Mistakes and Misconceptions

• Forgetting Δn = 0 – Many reactions have equal numbers of gas moles on both sides (e.g., Hâ‚‚ + Iâ‚‚ ⇌ 2HI, Δn = 0). Then Kp = Kc, but students often still multiply by RT.
• Using the wrong R – As noted, using 0.08206 when your pressures are in bar introduces an error of about 1.3%.
• Incorrect Δn – Δn is the difference in the stoichiometric coefficients of gaseous species only. Ignore solids and liquids. For example, in CaCO₃(s) ⇌ CaO(s) + COâ‚‚(g), Δn = 1 (only COâ‚‚ is gaseous).
• Temperature must be in kelvin – Always convert Celsius to K by adding 273.15. Using Celsius directly will give a wildly wrong result.
• Units of Kp – Kp has units of (pressure)^Δn. For Δn = 2, Kp has units atm². Many textbooks treat K as dimensionless, but when you do calculations with pressures, you must keep the units.

Limitations of the Kc–Kp Relationship

The formula Kp = Kc(RT)^Δn is derived from the ideal gas law, so it works well for gases at moderate pressures and temperatures where gases behave ideally. At very high pressures or low temperatures, real gases deviate from ideal behavior, and the relationship becomes approximate. In such cases, fugacity (a corrected pressure) is used instead of pressure, and the simple conversion no longer applies.

Also, the conversion assumes that the same reaction is being described – the stoichiometric coefficients in the balanced equation must match the equilibrium expression you are using. If you double the coefficients, both Kc and Kp change (they are raised to the power of the multiplier), and Δn also changes.

Frequently Asked Questions

What is the difference between Kc and Kp? Kc uses molar concentrations (mol/L) of all species, while Kp uses partial pressures of gases. For a reaction with gases, they are related by the equation above. If there are no gases, Kp is not defined.

Why does the temperature appear in the conversion? Because pressure and concentration are related through the ideal gas law, PV = nRT. The factor RT converts concentration (n/V) to pressure (P). At higher temperatures, the same concentration produces a higher pressure, so Kp becomes larger relative to Kc.

Can I convert Kp to Kc using the same formula? Yes, just rearrange: Kc = Kp / (RT)^Δn. The same rules for R and Δn apply.

What if Δn is negative? Then (RT)^Δn is a fraction (e.g., (RT)�¹ = 1/(RT)). For example, if Δn = −1, then Kp = Kc/(RT). The calculator handles negative exponents automatically.

Do solids and liquids affect Δn? No. Only gaseous species are included in Δn. Solids and liquids have negligible partial pressures and are omitted from the equilibrium expression.

Is Kp always larger than Kc? Not necessarily. If Δn > 0, Kp > Kc (because RT > 1 for typical temperatures). If Δn < 0, Kp < Kc. If Δn = 0, they are equal. At very low temperatures, RT could be less than 1, but in practice equilibrium reactions are usually studied above 200 K, where RT > 1.

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in kc | Kc value | mol/L | 0.5 | 0
in t | Temperature | K | 298 | 0
in dn | Δn (gas moles products − reactants) | - | 2 | -5
out kp | Kp value | (mol/L)^Δn·(R·T)^Δn | kc * (0.08206 * t)^dn | 4