IB Chemistry R2.3 R2.3.6

The Equilibrium Constant Kp

Expressing equilibrium using partial pressures for gaseous reactions.

Reactivity 2.3 HL Extension ⏱️ ~6 min revision
IB Understanding

Equilibrium Constants in Terms of Partial Pressure (Kp)

Equilibrium constants for gaseous reactions are expressed in terms of equilibrium partial pressures (\(K_{\text{p}}\)). Mole fractions and Dalton's law allow conversion between total pressure and individual partial pressures.

Key Definitions

TermDefinitionFormula
Mole fraction (χ)Ratio of moles of one gas to the total moles of all gases\(\chi_A = \frac{n_A}{n_{total}}\)
Partial pressureThe pressure a gas would exert if it alone occupied the container\(P_A = \chi_A \times P_{total}\)

The Kp Expression

For a gaseous equilibrium: \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\)

\[K_p = \frac{(P_C)^c(P_D)^d}{(P_A)^a(P_B)^b}\]
Examiner Trap

Gaseous Species Only in Kp Expressions

Only gaseous species (g) appear in \(K_{\text{p}}\) expressions. Pure solids (s) and pure liquids (l) have constant activities and are strictly omitted from \(K_{\text{p}}\).

Converting Between Kc and Kp

\[K_p = K_c(RT)^{\Delta n}\]

Where \(\Delta n = \text{total moles of gaseous products} - \text{total moles of gaseous reactants}\) and R = 8.314 J K⁻¹ mol⁻¹ (or 0.0821 atm L mol⁻¹ K⁻¹ depending on pressure units).

Worked Example: Calculating Kp

Worked Example 1

Converting Between Kc and Kp

Problem: For N₂(g) + 3H₂(g) ⇌ 2NH₃(g), Kc = 9.60 at 300 K. Calculate Kp.

Step 1: Find Δn = 2 − (1 + 3) = −2

Step 2: Apply the formula:

\(K_p = 9.60 \times (0.0821 \times 300)^{-2} = 9.60 \times (24.63)^{-2}\)

\(K_p = 9.60 \times 1.648 \times 10^{-3} = \textbf{0.0158}\)

Worked Example: Finding Mole Fractions & Partial Pressures

Worked Example 2

Calculating Kp from Partial Pressures

Problem: At equilibrium in the reaction N₂O₄(g) ⇌ 2NO₂(g) at 200 kPa total pressure, there are 0.40 mol N₂O₄ and 0.60 mol NO₂. Find Kp.

N₂O₄NO₂
Moles0.400.60
χ0.40/1.00 = 0.400.60/1.00 = 0.60
P (kPa)0.40 × 200 = 800.60 × 200 = 120

\(K_p = \frac{(120)^2}{80} = \frac{14400}{80} = \textbf{180 kPa}\)

Examiner Tip

Checking Mole Fractions and Δn Sign

Always check that your calculated mole fractions sum to exactly 1 (\(\sum \chi_i = 1\)). When using \(K_{\text{p}} = K_{\text{c}}(RT)^{\Delta n}\), verify \(\Delta n = \text{moles gas (products)} - \text{moles gas (reactants)}\).

AQA GCSE & IB Chemistry

Study this topic on the go

Get active recall flashcards, notes, and topic quizzes in ChemEasy, or build your revision schedule with ChemPlan IB.

See our apps