Kp and Partial Pressures
Kp and Partial Pressures
For gaseous equilibria, the equilibrium constant can be expressed in terms of partial pressures — this is Kp.
Partial Pressure
The partial pressure of a gas in a mixture is the pressure that gas would exert if it alone occupied the entire volume at the same temperature. It depends on the mole fraction of the gas:
p(A) = x(A) × P(total)
Where:
- p(A) is the partial pressure of gas A
- x(A) is the mole fraction of A = moles of A / total moles of gas
- P(total) is the total pressure of the mixture
Dalton's law: The total pressure of a mixture of gases is the sum of the partial pressures of all the gases present:
P(total) = p(A) + p(B) + p(C) + ...
Writing Kp Expressions
For the general equilibrium: aA(g) + bB(g) ⇌ cC(g) + dD(g)
Kp = [p(C)]^c × [p(D)]^d / [p(A)]^a × [p(B)]^b
Only gaseous species appear in the Kp expression. Solids and liquids are excluded.
Worked Example 1: Calculating Kp
Consider the equilibrium: N₂O₄(g) ⇌ 2NO₂(g)
At equilibrium at 400 K and 1.00 atm total pressure, the mole fraction of NO₂ is 0.40.
Step 1: Find mole fractions
- x(NO₂) = 0.40
- x(N₂O₄) = 1 − 0.40 = 0.60
Step 2: Find partial pressures
- p(NO₂) = 0.40 × 1.00 = 0.40 atm
- p(N₂O₄) = 0.60 × 1.00 = 0.60 atm
Step 3: Calculate Kp
- Kp = [p(NO₂)]² / p(N₂O₄) = (0.40)² / 0.60 = 0.16 / 0.60 = 0.267 atm
Worked Example 2: ICE Table Method
2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
A vessel initially contains 4.0 mol SO₂ and 3.0 mol O₂ at 700 K and 5.0 atm total pressure. At equilibrium, 3.0 mol SO₃ is present.
ICE table (moles):
| SO₂ | O₂ | SO₃ | |
|---|---|---|---|
| Initial | 4.0 | 3.0 | 0 |
| Change | −3.0 | −1.5 | +3.0 |
| Equilibrium | 1.0 | 1.5 | 3.0 |
Total moles at equilibrium = 1.0 + 1.5 + 3.0 = 5.5
Mole fractions:
- x(SO₂) = 1.0/5.5 = 0.182
- x(O₂) = 1.5/5.5 = 0.273
- x(SO₃) = 3.0/5.5 = 0.545
Partial pressures (× 5.0 atm):
- p(SO₂) = 0.909 atm
- p(O₂) = 1.364 atm
- p(SO₃) = 2.727 atm
Kp = [p(SO₃)]² / [p(SO₂)]² × p(O₂)
Kp = (2.727)² / [(0.909)² × 1.364] = 7.437 / 1.127 = 6.60 atm⁻¹
Units of Kp
The units of Kp depend on the expression. Work them out by substituting pressure units:
- If the powers of pressure in the numerator and denominator are equal, Kp has no units
- If there are more moles of gas in products, the unit involves pressure raised to a positive power (e.g. atm, kPa)
- If there are more moles of gas in reactants, the unit involves pressure raised to a negative power (e.g. atm⁻¹, kPa⁻¹)
Effect of Changing Conditions on Kp
Temperature:
- Kp changes with temperature (it is a true equilibrium constant only at a fixed temperature)
- For an exothermic forward reaction: increasing T shifts equilibrium left → Kp decreases
- For an endothermic forward reaction: increasing T shifts equilibrium right → Kp increases
Pressure:
- Kp does not change with pressure (at constant temperature)
- However, individual partial pressures and the equilibrium position do change
- Increasing total pressure shifts equilibrium towards the side with fewer moles of gas (Le Chatelier), but Kp stays the same
Catalyst:
- A catalyst does not change Kp — it speeds up both forward and reverse reactions equally
- Equilibrium is reached faster but at the same position
Kp vs Kc
| Feature | Kc | Kp |
|---|---|---|
| Uses | Concentrations (mol dm⁻³) | Partial pressures |
| Applies to | Any equilibrium in solution or gas phase | Gaseous equilibria only |
| Related by | Kp = Kc × (RT)^Δn | where Δn = moles gas products − moles gas reactants |
When Δn = 0, Kp = Kc (numerically, though units differ).
Heterogeneous Equilibria
In a heterogeneous equilibrium, reactants and products are in different phases. Only gaseous species appear in Kp:
CaCO₃(s) ⇌ CaO(s) + CO₂(g) Kp = p(CO₂)
The partial pressure of CO₂ alone determines whether equilibrium is established.
Exam Tips
- Set out ICE tables clearly — examiners follow your working through these
- Always state and derive the units of Kp
- When the question gives kPa, work in kPa throughout — do not mix with atm
- Remember: changing pressure changes the equilibrium position but not Kp (at constant T)