Kp and Kc Conversion

Introduction

Data tables sometimes give an equilibrium constant in pressures when your problem is set in concentrations, or the other way round. The two constants describe the same equilibrium, but they have different numerical values whenever the number of gas molecules changes. The conversion is short, yet it is a favourite source of errors: the wrong value of R, the wrong Δn, or a Celsius temperature. This page derives the relation and shows how to apply it cleanly.

Core explanation

Derivation. For an ideal gas, pᵢ = (nᵢ/V)RT = cᵢRT, where cᵢ is the molar concentration. For a general gas reaction aA + bB ⇌ cC + dD:

Kp = p C^c p D^d / (p A^a p B^b) = [C]^c[D]^d (RT)^(c+d) / ([A]^a[B]^b (RT)^(a+b))

so Kp = Kc (RT)^Δn, where Δn = (c + d) − (a + b), counting gases only.

Units of R must match. If concentrations are in mol dm⁻³ and pressures in bar, use R = 0.08314 bar dm³ mol⁻¹ K⁻¹. If concentrations are in mol m⁻³ and pressures in Pa, use R = 8.314 Pa m³ mol⁻¹ K⁻¹ (identical to J mol⁻¹ K⁻¹). Mixing R = 8.314 with bar and dm³ gives an answer wrong by a factor of 100 for each power of Δn. Temperature must always be in kelvin.

Three cases.

- Δn = 0 (for example H₂ + I₂ ⇌ 2HI): Kp = Kc, and both are dimensionless. - Δn > 0 (for example N₂O₄ ⇌ 2NO₂): Kp > Kc whenever RT > 1 in the chosen units. With Kc = 4.6 × 10⁻³ mol dm⁻³ at 298 K, RT = 0.08314 × 298 = 24.8 bar dm³ mol⁻¹, so Kp = 4.6 × 10⁻³ × 24.8 = 0.11 bar. - Δn < 0 (for example 2SO₂ + O₂ ⇌ 2SO₃, Δn = −1): Kp = Kc / RT. A commonly quoted value is Kc ≈ 2.8 × 10² dm³ mol⁻¹ at 1000 K; RT = 83.1 bar dm³ mol⁻¹, so Kp ≈ 280 / 83.1 = 3.4 bar⁻¹.

Heterogeneous equilibria. Pure solids and liquids do not appear in either expression, and they are not counted in Δn. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Δn = +1, Kc = [CO₂] and Kp = p(CO₂).

Converting units at the end. Because the conversion introduces (RT)^Δn, the units of the result change too: Kp has units of bar^Δn and Kc of (mol dm⁻³)^Δn. Writing these units out and checking that they cancel properly is the fastest way to catch a slip.

Formulae

Kp = Kc(RT)^Δn and Kc = Kp(RT)^(−Δn). R = 0.08314 bar dm³ mol⁻¹ K⁻¹ (for bar and mol dm⁻³) or 8.314 J mol⁻¹ K⁻¹ (for Pa and mol m⁻³). T in K. Δn counts gaseous species only.

Step-by-step reasoning

1. Balance the equation and mark the states. 2. Count Δn using gases only. 3. Pick R to match the pressure and concentration units. 4. Compute RT with T in kelvin. 5. Multiply or divide K by (RT)^ Δn as required and assign units.

Visual explanation

Think of the factor RT as an exchange rate between two currencies: concentration and pressure. Each gaseous product brings one factor of RT into the numerator, and each gaseous reactant one into the denominator. The net number of factors that survive is Δn, just as a net count of coins converted determines the total exchange fee.

Real-world analogy

Converting a recipe from cups to millilitres multiplies each ingredient by the same factor. If a ratio has more ingredients on top than on the bottom, the conversion factor survives once for each extra ingredient. Kp and Kc are two "recipe ratios" in different units, and Δn tells you how many conversion factors are left over.

Real-world example

Thermodynamic tables give standard Gibbs energies from which dimensionless pressure-based constants are calculated. Reactor models for gas-phase processes such as sulfuric-acid manufacture then often need concentration-based constants for rate expressions, so engineers routinely apply this conversion at the operating temperature.

Why?

Why does RT appear? Because the ideal-gas law links pressure to concentration by p = cRT. Every gas term in the expression is converted by this factor, and those factors cancel except for the excess, which is the change in the number of gas moles.

Common misconception

"Δn is the total number of moles on the product side." It is the difference between gaseous products and gaseous reactants. Including solids, liquids or dissolved species, or forgetting to subtract the reactant side, gives the wrong power.

Worked example

Question: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g) at 500 K, suppose Kp = 3.6 × 10⁻² bar⁻². Calculate Kc.

Reasoning: Δn = 2 − 4 = −2. RT = 0.08314 × 500 = 41.57 bar dm³ mol⁻¹. Kc = Kp(RT)^(−Δn) = Kp(RT)² = 3.6 × 10⁻² × (41.57)² = 3.6 × 10⁻² × 1728 = 62.

Answer: Kc ≈ 62 dm⁶ mol⁻² (units: bar⁻² × bar² dm⁶ mol⁻²).

Quick check

1. What is Δn for 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g), and how are Kp and Kc related? Answer: Δn = 10 − 9 = +1, so Kp = Kc × RT.

Exam focus

Show the value and units of R you are using, calculate Δn explicitly, and convert temperature to kelvin. A correct final answer without units usually loses a mark. Look out for questions that include a solid or liquid water to test whether you count only gases.

Advanced insight

Thermodynamic constants are dimensionless: K° = Kp/(p°)^Δn with p° = 1 bar, and Kc° = Kc/(c°)^Δn with c° = 1 mol dm⁻³. The exact relation is Kp° = Kc°(c°RT/p°)^Δn. Using ΔrG° = −RT ln K° requires the constant referenced to the same standard state as the tabulated Gibbs energies, which for gases is 1 bar.

Summary

For ideal gases, p = cRT gives Kp = Kc(RT)^Δn, where Δn counts gaseous species only. When Δn = 0 the constants are equal; otherwise the value of R must match the units (0.08314 bar dm³ mol⁻¹ K⁻¹ for bar and mol dm⁻³) and T must be in kelvin. Units of the constants change as bar^Δn and (mol dm⁻³)^Δn.

Practice questions

1. For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kp = 1.00 bar at about 1170 K. Calculate Kc. Answer: Δn = +1 (only CO₂ counts); RT = 0.08314 × 1170 = 97.3, so Kc = 1.00 / 97.3 = 1.03 × 10⁻² mol dm⁻³. 2. Why is Kp equal to Kc for H₂(g) + Cl₂(g) ⇌ 2HCl(g)? Answer: Δn = 2 − 2 = 0, so (RT)^0 = 1 and the two constants are identical and dimensionless. 3. A student converts Kc for N₂O₄ ⇌ 2NO₂ using R = 8.314 while working in bar and mol dm⁻³. By what factor is the answer wrong? Answer: Δn = 1, so the result is 100 times too large, because 8.314 is 100 times the correct 0.08314 for those units. 4. For 2SO₂ + O₂ ⇌ 2SO₃, Kp = 3.4 bar⁻¹ at 1000 K. Find Kc. Answer: Δn = −1, so Kc = Kp × RT = 3.4 × 83.1 ≈ 2.8 × 10² dm³ mol⁻¹.