Kp vs Kc

When a reversible chemical reaction reaches dynamic equilibrium, the ratio of product concentrations to reactant concentrations remains constant at a given temperature. Chemists express this ratio in two equivalent but distinct ways: the equilibrium constant in terms of molar concentrations, K_c, and the equilibrium constant in terms of partial pressures, K_p. Both constants describe the same equilibrium position, but they use different measurable quantities — molarity (mol/L) for K_c and partial pressure (usually in atm or bar) for K_p. The two constants are related through the ideal gas law. Because pressure and concentration are connected by P = (n/V)RT = cRT, any change in the number of moles of gas between reactants and products will cause K_p and K_c to differ by a factor involving the gas constant R and the absolute temperature T. This relationship is captured in the central equation of this topic: K_p = K_c(RT)^{Δn_g}, where Δn_g is the change in moles of gas in the balanced equation. Understanding when to use K_p versus K_c is essential for solving equilibrium problems. For reactions involving only solids and liquids, K_p and K_c are essentially not applicable in the pressure form, so K_c is preferred. For gas-phase reactions where Δn_g = 0, K_p equals K_c exactly. Knowing the units (or conventional dimensionlessness) of K also matters: in rigorous thermodynamics, equilibrium constants are dimensionless because each concentration or pressure is divided by a standard reference value (1 mol/L or 1 bar), but at the high-school level it is common to quote units consistent with the expression used.

Key takeaways
  • Write the General Equilibrium Expression
  • Apply the Ideal Gas Law to Each Species
  • Substitute into the K_p Expression
  • Define Δn_g and Write the Final Relation
  • Three Key Cases for Δn_g
  • A Note on Units and Dimensionless K

Write the General Equilibrium Expression

For a general gas-phase reaction a A(g) + b B(g) ⇌ c C(g) + d D(g), the two equilibrium constants are defined as: K_c = [C]^c[D]^d / ([A]^a[B]^b), using molar concentrations in mol/L, and K_p = (P_C)^c(P_D)^d / ((P_A)^a(P_B)^b), using partial pressures.

Apply the Ideal Gas Law to Each Species

From the ideal gas law PV = nRT, we get P = (n/V)RT = [X]RT for any gaseous species X, where [X] = n/V is the molar concentration. Therefore the partial pressure of each gas equals its molar concentration multiplied by RT: P_A = [A]RT, P_B = [B]RT, P_C = [C]RT, P_D = [D]RT.

Substitute into the K_p Expression

Replacing each partial pressure in the K_p expression with [X]RT gives: K_p = ([C]RT)^c([D]RT)^d / (([A]RT)^a([B]RT)^b). Expanding the powers of RT: K_p = [C]^c[D]^d(RT)^c(RT)^d / ([A]^a[B]^b(RT)^a(RT)^b) = K_c × (RT)^(c+d) / (RT)^(a+b) = K_c × (RT)^((c+d)-(a+b)).

Define Δn_g and Write the Final Relation

Define Δn_g = (moles of gaseous products) − (moles of gaseous reactants) = (c + d) − (a + b). The derivation then gives the elegant result: K_p = K_c(RT)^{Δn_g}. Note that R = 0.0821 L·atm/(mol·K) when pressures are in atm, or R = 0.08314 L·bar/(mol·K) when pressures are in bar. Always use T in Kelvin.

Three Key Cases for Δn_g

Case 1 — Δn_g = 0: K_p = K_c × (RT)^0 = K_c × 1 = K_c. Example: \ce{H2(g) + I2(g) <=> 2HI(g)}, Δn_g = 2 − 2 = 0. Case 2 — Δn_g > 0: K_p > K_c (more gas moles on product side increases pressure relative to concentration). Example: \ce{PCl5(g) <=> PCl3(g) + Cl2(g)}, Δn_g = 2 − 1 = +1, so K_p = K_c(RT). Case 3 — Δn_g < 0: K_p < K_c (fewer gas moles produced means pressure ratio is smaller). Example: \ce{N2(g) + 3H2(g) <=> 2NH3(g)}, Δn_g = 2 − 4 = −2, so K_p = K_c(RT)^{-2} = K_c/(RT)^2.

A Note on Units and Dimensionless K

Strictly speaking, equilibrium constants in thermodynamics are dimensionless: each concentration is divided by the standard concentration c° = 1 mol/L, and each pressure by the standard pressure p° = 1 bar (or 1 atm in older texts). This means K_p and K_c are actually pure numbers, and Δn_g carries no unit baggage because (RT/p°) effectively has units that cancel. At the high-school level, it is acceptable to write units on K_c as (mol/L)^{Δn_g} or on K_p as (atm)^{Δn_g}, as long as you are consistent throughout a calculation.

The Key Equation

Kp=Kc(RT)ΔngK_p = K_c(RT)^{\Delta n_g}
  • K_p — Pressure equilibrium constant (dimensionless (or atm^{Δn_g} in practice)): Equilibrium constant expressed using partial pressures of gaseous species.
  • K_c — Concentration equilibrium constant (dimensionless (or (mol/L)^{Δn_g} in practice)): Equilibrium constant expressed using molar concentrations of species in solution or gas phase.
  • R — Ideal gas constant (L·atm/(mol·K) or L·bar/(mol·K)): Universal gas constant; use R = 0.0821 L·atm/(mol·K) when pressures are in atm.
  • T — Absolute temperature (K (Kelvin)): Temperature must always be expressed in Kelvin in this equation.
  • \Delta n_g — Change in moles of gas (dimensionless (mol of gas)): Δn_g = (total moles of gaseous products) − (total moles of gaseous reactants) from the balanced equation.

Worked Example

Problem. At 500 K, the equilibrium constant K_c for the reaction \ce{N2(g) + 3H2(g) <=> 2NH3(g)} is 6.0 × 10^{-2} (mol/L)^{-2}. Calculate K_p at this temperature. (Use R = 0.0821 L·atm/(mol·K).)

Kp=Kc(RT)2=6.0×10−2(41.05)2≈3.6×10−5K_p = \frac{K_c}{(RT)^{2}} = \frac{6.0 \times 10^{-2}}{(41.05)^{2}} \approx 3.6 \times 10^{-5}

Answer. K_p ≈ 3.6 × 10^{-5} at 500 K

Figure
Diagram
K_p vs K_c: Effect of Δn_g on the RelationshipTemperature T (K)K valueΔnᵍ > 0Δnᵍ = 0Δnᵍ < 0Low THigh TKey EquationsΔnᵍ > 0 : Kₚ > Kc (Kₚ rises with T)Δnᵍ = 0 : Kₚ = Kc (no T dependence)Δnᵍ < 0 : Kₚ < Kc (Kₚ falls with T)H₂(g) + I₂(g) ⇌ 2HI(g)PCl₅ ⇌ PCl₃ + Cl₂N₂ + 3H₂ ⇌ 2NH₃0
Origins · 1 minJacobus Henricus van 't Hoff and how this was discovered
JH
Jacobus Henricus van 't Hoff
1852–1911 · Dutch

Van 't Hoff was the first winner of the Nobel Prize in Chemistry (1901). He developed the mathematical framework for chemical equilibrium and thermodynamics, including the van 't Hoff equation d(ln K)/dT = ΔH°/RT², which describes how equilibrium constants change with temperature. His early work on osmotic pressure also established the connection between gas laws and solution behaviour, directly underpinning the derivation of K_p = K_c(RT)^{Δn_g}. Van 't Hoff initially struggled to get his landmark 1874 paper on tetrahedral carbon accepted — a senior chemist publicly dismissed it as 'fantastic nonsense' from someone spending too much time riding horses. The paper went on to found the field of stereochemistry.

  • Writing the Equilibrium Constant Expression (K_c) — K_c is one of the two quantities in the K_p = K_c(RT)^{Δn_g} equation; understanding how to construct K_c from a balanced equation is a prerequisite.
  • Le Chatelier's Principle — Changes in pressure shift gas-phase equilibria; knowing the difference between K_p and K_c helps predict how pressure changes affect equilibrium position.
  • The Ideal Gas Law (PV = nRT) — The derivation of K_p = K_c(RT)^{Δn_g} is based entirely on substituting P = [X]RT from the ideal gas law into the K_p expression.
  • The Reaction Quotient Q — Q can also be expressed in pressure (Q_p) or concentration (Q_c) terms, and the same relationship Q_p = Q_c(RT)^{Δn_g} applies when predicting the direction a reaction will shift.
  • Gibbs Free Energy and Equilibrium (ΔG° = −RT ln K) — The thermodynamic equilibrium constant used in ΔG° = −RT ln K is the dimensionless form of K_p referenced to standard pressure, connecting K_p directly to the spontaneity of reactions.