Edexcel A-level Chemistry (9CH0) · Topic 11: Equilibrium II
Mini-Lesson
Equilibrium II
This mini-lesson covers the whole of Edexcel Topic 11: mole fractions and partial pressures, writing and calculating Kp for homogeneous gaseous equilibria (including its units), the effect of temperature on the value of K, and exactly why catalysts and pressure never change it.
Work through each screen, answer the questions as you go (most are calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Partial pressure
Mole fractions and partial pressures
In a mixture of gases, the partial pressure of a gas is the pressure that gas would exert if it alone occupied the whole container. The partial pressures of all the gases add up to the total pressure (Dalton's law).
mole fraction, x = n(gas) ÷ n(total gas)the mole fractions of all the gases in a mixture always add up to 1
partial pressure, p = x × P(total)and so the partial pressures always add up to P(total)
Worked example
A mixture contains 2.0 mol N₂ and 6.0 mol H₂ at a total pressure of 400 kPa. Total = 8.0 mol.
Always check. Your mole fractions must sum to 1 and your partial pressures must sum to the total. This catches almost every arithmetic slip.
Calculate
Your turn — mole fraction
1An equilibrium mixture contains 0.400 mol N₂, 1.20 mol H₂ and 0.400 mol NH₃. Calculate the mole fraction of H₂.
(no units)
Hint: total moles of gas = 0.400 + 1.20 + 0.400 = 2.00 mol. x(H₂) = 1.20 ÷ 2.00.
Calculate
Your turn — partial pressure
2That same mixture is at a total pressure of 250 kPa. Calculate the partial pressure of H₂ in kPa.
kPa
Hint: p(H₂) = x(H₂) × P(total) = 0.600 × 250.
Kp
The Kp expression
Kp is the equilibrium constant written in terms of partial pressures instead of concentrations. It is used for homogeneous gaseous equilibria — every species must be a gas.
aA(g) + bB(g) ⇌ cC(g) + dD(g) Kp = p(C)c × p(D)d ÷ (p(A)a × p(B)b)same shape as Kc: products on top, each raised to the power of its coefficient
Only gases appear. Pure solids and pure liquids are left out entirely — their "pressure" does not vary.
Kp is constant at a given temperature, exactly like Kc.
The units depend on the equation. Substitute the pressure unit (kPa or atm) and cancel.
Many Kp questions give you a starting amount and a percentage that dissociates. Build the equilibrium moles first — the total number of moles usually changes, and that is exactly the point.
Worked example — dinitrogen tetroxide
1.00 mol of N₂O₄ is heated and 20.0% dissociates: N₂O₄(g) ⇌ 2NO₂(g). Total pressure = 100 kPa.
Watch the total. Here 1.00 mol of gas became 1.200 mol — dissociation makes more gas particles, so the total moles is not the starting value.
Calculate
Your turn — degree of dissociation
71.00 mol of PCl₅ is heated in a sealed vessel and 40.0% of it dissociates: PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). The total pressure at equilibrium is 200 kPa. Calculate Kp in kPa (3 s.f.).
8For H₂(g) + I₂(g) ⇌ 2HI(g), Kp = 64.0 at a given temperature. In an equilibrium mixture, p(H₂) = p(I₂) = 20.0 kPa. Calculate p(HI) in kPa.
kPa
Hint: rearrange — p(HI)² = Kp × p(H₂) × p(I₂) = 64.0 × 20.0 × 20.0 = 25 600. Now take the square root.
Temperature
Temperature is the only thing that changes K
The value of Kp (like Kc) depends on temperature and nothing else. Use Le Chatelier to work out which way it moves:
Exothermic forward reaction (ΔH negative): raising T shifts the equilibrium left, so there is proportionally less product at the new equilibrium and Kp decreases. Lowering T increases Kp.
Endothermic forward reaction (ΔH positive): raising T shifts the equilibrium right, so Kp increases.
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJ mol⁻¹heat it and Kpfalls — this is exactly why the Contact process has to compromise on temperature
Say it precisely in the exam: "The forward reaction is exothermic, so increasing the temperature shifts the position of equilibrium in the endothermic (reverse) direction. The partial pressure of SO₃ falls and those of SO₂ and O₂ rise, so Kp decreases."
Pressure & catalysts
Why pressure and catalysts never change K
This is the single most tested idea in Topic 11 — and the one students most often get wrong.
Increasing the total pressure increases every partial pressure at the instant of the change. For N₂ + 3H₂ ⇌ 2NH₃ the denominator (with four powers of pressure) grows faster than the numerator (with two), so the quotient momentarily falls below Kp. The system responds by shifting right until the quotient equals Kp once more. The position moves; the constant does not.
A catalyst lowers the activation energy of the forward and reverse reactions by the same amount, so it multiplies both rates equally. The rates were already equal at equilibrium, so they remain equal — nothing shifts, and Kp is untouched. A catalyst only makes the system reach equilibrium sooner.
Adding an inert gas at constant volume raises the total pressure but leaves the partial pressure of every reacting gas unchanged — so nothing shifts and Kp is unchanged.
The one-line summary: concentration, pressure and catalysts can move the position of an equilibrium, but only a change in temperature can move the value of K.
Quick check
Pressure and Kp
9The total pressure on the equilibrium N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is doubled at constant temperature. What happens to Kp?
Quick check
Temperature and Kp
10For N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57 kJ mol⁻¹. The temperature is increased. What happens to Kp?
Quick check
A catalyst
11V₂O₅ is added to the Contact process equilibrium. What is the effect on Kp and on the equilibrium yield of SO₃?
Quick check
What Kp tells you
12Two gaseous equilibria are compared at the same temperature. Reaction A has Kp = 1.6 × 10⁵; reaction B has Kp = 3.0 × 10⁻⁴ (both dimensionless). What does this tell you?
Sort it
Does it change Kp, the position, or nothing?
The system is 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ mol⁻¹, at equilibrium in a closed vessel. Tap a change, then tap the bin it belongs in.
🔑 Changes the value of Kp
↔️ Moves the position, but Kp is unchanged
⏸️ Changes neither
Match it
Match the term to its meaning
Tap an item on the left, then its partner on the right.
Term
Meaning
Recap
The big ideas to know
Mole fraction: x = n(gas) ÷ n(total gas) — the mole fractions always sum to 1
Partial pressure: p = x × P(total) — the partial pressures always sum to P(total)
Kp: products over reactants, each partial pressure raised to the power of its coefficient; gases only
Units: substitute kPa and cancel — kPa⁻² for the Haber equilibrium, kPa⁻¹ for the Contact, none for H₂ + I₂ ⇌ 2HI