📘 IB Understanding
Equilibrium constants for reactions involving gases can be expressed in terms of partial pressures (Kp). The relationship between Kc and Kp depends on the change in the number of moles of gas, Δn.
Key Definitions
| Term | Definition | Formula |
|---|---|---|
| Mole fraction (χ) | Ratio of moles of one gas to the total moles of all gases | \(\chi_A = \frac{n_A}{n_{total}}\) |
| Partial pressure | The 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: \(aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g)\)
💡 Important
Only gaseous species (g) appear in Kp expressions. Solids, liquids, and aqueous species are omitted.
Converting Between Kc and Kp
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
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
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.
| N₂O₄ | NO₂ | |
|---|---|---|
| Moles | 0.40 | 0.60 |
| χ | 0.40/1.00 = 0.40 | 0.60/1.00 = 0.60 |
| P (kPa) | 0.40 × 200 = 80 | 0.60 × 200 = 120 |
\(K_p = \frac{(120)^2}{80} = \frac{14400}{80} = \textbf{180 kPa}\)
📋 Exam Tips
- Check that all mole fractions sum to 1 and all partial pressures sum to Ptotal
- Be careful with the sign of Δn — a negative Δn makes Kp < Kc
- If Δn = 0, then Kp = Kc (the pressure units cancel)