Gas-Phase Equilibrium Constant: Kp vs Kc

For gas-phase reactions, the equilibrium constant can be written with partial pressures (Kp) or with concentrations (Kc), and the two are related by Kp = Kc(RT)Δn, where Δn is the change in the number of moles of gas. In brief: for N2O4 ⇌ 2NO2 (Δn = +1) with Kc = 0.0060 at 298 K, Kp = 0.0060 × 0.08206 × 298 = 0.147.

Gas concentrations are often given in terms of pressure, so the equilibrium expression is written with partial pressures. The two constants have different numerical values, except when the number of gas moles does not change. This article defines Kp, derives the relationship, tells you when Kp and Kc are equal and works through the conversion in both directions.

What is Kp?

For the gas-phase reaction aA(g) + bB(g) ⇌ cC(g) + dD(g), the equilibrium constant in terms of partial pressures is:

Kp = (PC)c (PD)d / ((PA)a (PB)b)

The partial pressures are those at equilibrium, usually in atmospheres or bar. As with Kc, pure solids and liquids are left out, so only gases appear. For example, for 2SO2 + O2 ⇌ 2SO3, Kp = PSO32 / (PSO22 · PO2).

How are Kp and Kc related?

For an ideal gas, PV = nRT, so the partial pressure is P = (n/V)RT = cRT, where c is the molar concentration. Replacing each pressure in the Kp expression by cRT and collecting the powers of RT gives:

Kp = Kc (RT)Δn

where Δn = (moles of gaseous products) − (moles of gaseous reactants), counting only gases. R = 0.08206 L·atm/(mol·K) when pressure is in atmospheres and concentrations in mol/L, and T is in kelvins. Rearranged: Kc = Kp(RT)−Δn.

N2O4 ⇌ 2NO2 ⇌ Δn = +1 N2 + 3H2 ⇌ 2NH3 ⇌ Δn = −2

When are Kp and Kc equal?

The change in moles of gas and the relation between Kp and Kc
Reaction Δn Relation
H2(g) + I2(g) ⇌ 2HI(g)2 − 2 = 0Kp = Kc
PCl5(g) ⇌ PCl3(g) + Cl2(g)2 − 1 = +1Kp = Kc(RT)
2SO2(g) + O2(g) ⇌ 2SO3(g)2 − 3 = −1Kp = Kc/(RT)
N2(g) + 3H2(g) ⇌ 2NH3(g)2 − 4 = −2Kp = Kc/(RT)2
CaCO3(s) ⇌ CaO(s) + CO2(g)1 − 0 = +1Kp = Kc(RT)

The two constants are the same when Δn = 0. In the last example, only the gas counts: solids are not included in Δn.

Worked examples

Example 1: converting Kc to Kp

For N2O4(g) ⇌ 2NO2(g), Kc = 0.0060 at 298 K. Find Kp.

  1. Δn = 2 − 1 = +1.
  2. RT = 0.08206 × 298 = 24.45 L·atm/mol.
  3. Kp = 0.0060 × 24.45 = 0.147.

Example 2: converting Kp to Kc

For N2 + 3H2 ⇌ 2NH3, Kp = 1.45 × 10−5 at 500 °C (773 K). Find Kc.

  1. Δn = −2, so Kc = Kp(RT)2.
  2. RT = 0.08206 × 773 = 63.43.
  3. Kc = 1.45 × 10−5 × (63.43)2 = 1.45 × 10−5 × 4023 = 0.058.

Example 3: finding Kp from pressures

For PCl5 ⇌ PCl3 + Cl2 at 500 K, the equilibrium partial pressures are P(PCl5) = 0.80 atm, P(PCl3) = 0.20 atm and P(Cl2) = 0.20 atm.

  1. Kp = (0.20)(0.20) / 0.80 = 0.050.
  2. RT = 0.08206 × 500 = 41.03.
  3. Kc = Kp/(RT) = 0.050 / 41.03 = 1.2 × 10−3.

Example 4: no difference when Δn = 0

For H2 + I2 ⇌ 2HI, Kc = 54.3 at 430 °C, and so Kp is also 54.3, since (RT)0 = 1.

Which units and standard states are used?

Strictly, the thermodynamic equilibrium constant K is dimensionless: pressures are divided by the standard pressure (1 bar) and concentrations by the standard concentration (1 mol/L). In textbooks Kp is often given a unit such as atmΔn, or none. For conversions, the value of R should match the pressure unit: if pressure is in bar, use R = 0.08314 L·bar/(mol·K), and if it is in kPa, use 8.314 L·kPa/(mol·K). If the pressure unit is atm, the value 0.08206 gives Kp in atm. The difference between atm and bar is small (1 atm = 1.01325 bar).

Where does the relation come from?

The use of partial pressures in equilibrium expressions follows from the ideal gas law and the work of Guldberg and Waage on the law of mass action in the 1860s. Pressures are easier to measure than concentrations for gases, so partial pressures became the natural choice in studies of gas equilibria. The ammonia equilibrium, studied by Fritz Haber and Walther Nernst in the years around 1905–1910 as a step toward the process that bears Haber’s name, is a typical example of such a gas-phase system.

Which mistakes are common?