Edexcel A-level Chemistry (9CH0) · Topic 10: Equilibrium I
Mini-Lesson
Equilibrium I
This mini-lesson covers the whole of Edexcel Topic 10: dynamic equilibrium, Le Chatelier's principle (concentration, pressure and temperature), why a catalyst does nothing to the position, the Kc expression and how to calculate it, and the industrial compromise behind the Haber and Contact processes.
Work through each screen, answer the questions as you go (several are Kc calculations) and collect ⭐ stars. Every calculation is worked through for you first. Press Start when you're ready.
Dynamic equilibrium
What "dynamic equilibrium" means
A reversible reaction in a closed system reaches dynamic equilibrium when:
the rate of the forward reaction equals the rate of the reverse reaction; and
the concentrations of all species remain constant.
It is dynamic, not static: both reactions carry on at exactly the same rate, so nothing appears to change even though molecules are constantly reacting in both directions.
Three traps to avoid. (1) "Constant" does not mean "equal" — the concentrations of reactants and products are almost never the same. (2) The system must be closed (nothing enters or leaves); an open beaker cannot reach equilibrium if a gas escapes. (3) Equilibrium can be reached from either direction and gives the same position at a given temperature.
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)the ⇌ sign is the whole point: the reaction goes both ways
Quick check
Spot the true statement
1Which statement about a system at dynamic equilibrium is correct?
Le Chatelier
Le Chatelier's principle
If a change is made to a system at equilibrium, the position of equilibrium shifts so as to oppose the changeit minimises the change — it never completely cancels it
Applied to the three things you can change:
Concentration. Increase the concentration of a reactant → the system shifts right to use it up. Remove a product as it forms → the system shifts right to replace it.
Pressure (gases only). Increase the pressure → the system shifts to the side with fewer moles of gas, to reduce the pressure. If both sides have the same number of gas moles, pressure has no effect on the position.
Temperature. Increase the temperature → the system shifts in the endothermic direction, to absorb the extra heat. Decrease it → it shifts in the exothermic direction.
The one that also changes K: only temperature changes the value of the equilibrium constant. Concentration and pressure shift the position but leave Kc unchanged — the system moves to restore the same value of Kc.
Le Chatelier
Working an example: the Haber process
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹4 moles of gas on the left, 2 moles of gas on the right; the forward reaction is exothermic
Increase the pressure → shifts right (4 mol → 2 mol reduces the pressure). Yield of NH₃ up.
Increase the temperature → shifts left (the reverse reaction is endothermic). Yield of NH₃ down, and Kc falls.
Remove NH₃ by condensing it out → shifts right. Yield up.
Add an iron catalyst → no shift at all. Equilibrium is simply reached faster.
Why a catalyst cannot shift the position: it lowers the activation energy of the forward and the reverse reaction by the same amount, so both rates increase equally. Equal rates before, equal rates after — the position, and Kc, are untouched.
Quick check
Pressure and the position
2For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), what happens to the equilibrium yield of SO₃ when the pressure is increased at constant temperature?
Quick check
Temperature and the position
3For N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57 kJ mol⁻¹. The sealed tube is placed in hot water. What happens?
Kc
The equilibrium constant, Kc
For a general homogeneous equilibrium in solution or in the gas phase:
aA + bB ⇌ cC + dD Kc = [C]c[D]d ÷ ([A]a[B]b)products on top, reactants underneath; each concentration is raised to the power of its coefficient
Square brackets mean equilibrium concentration in mol dm⁻³ — not starting concentration.
Kc is constant at a given temperature and changes only when the temperature changes.
A large Kc (≫ 1) means the equilibrium lies well to the right: mostly products. A small Kc (≪ 1) means mostly reactants.
The units depend on the equation — work them out each time by substituting mol dm⁻³ and cancelling.
Units worked out: for N₂O₄ ⇌ 2NO₂, Kc = [NO₂]² ÷ [N₂O₄], so the units are (mol dm⁻³)² ÷ (mol dm⁻³) = mol dm⁻³. For H₂ + I₂ ⇌ 2HI the units cancel completely, so Kc has no units.
Method for "moles" questions: (1) build an ICE table — Initial, Change, Equilibrium moles; (2) divide each equilibrium amount by the volume to get concentrations; (3) substitute. If the total number of moles of gas is the same on both sides, the volume cancels and you may use moles directly.
Calculate
Your turn — Kc from concentrations
4For H₂(g) + I₂(g) ⇌ 2HI(g), the equilibrium concentrations are [H₂] = 0.0060, [I₂] = 0.0060 and [HI] = 0.0440 mol dm⁻³. Calculate Kc to 3 significant figures.
Never put moles into Kc without dividing by the volume — unless the moles of gas are equal on both sides, in which case the volumes genuinely cancel.
Calculate
Your turn — Kc from equilibrium moles
5In a 5.00 dm³ vessel at equilibrium there are 0.50 mol PCl₅, 1.50 mol PCl₃ and 1.50 mol Cl₂. PCl₅(g) ⇌ PCl₃(g) + Cl₂(g). Calculate Kc in mol dm⁻³.
mol dm⁻³
Hint: divide by 5.00 first — [PCl₅] = 0.10, [PCl₃] = 0.30, [Cl₂] = 0.30. Then Kc = (0.30 × 0.30) ÷ 0.10.
Calculate
Your turn — the volume cancels
61.00 mol of ethanoic acid and 1.00 mol of ethanol are mixed and reach equilibrium; 0.33 mol of the acid remains. CH₃COOH + C₂H₅OH ⇌ CH₃COOC₂H₅ + H₂O. Calculate Kc to 3 s.f.
(no units)
Hint: 0.67 mol of acid reacted, so 0.67 mol of ester and 0.67 mol of water formed, and 0.33 mol of ethanol remains. There are 2 moles each side, so the volume cancels: Kc = (0.67 × 0.67) ÷ (0.33 × 0.33).
Calculate
Your turn — working backwards from Kc
7For H₂(g) + I₂(g) ⇌ 2HI(g), Kc = 54.0 at 700 K. At equilibrium [H₂] = [I₂] = 0.0100 mol dm⁻³. Calculate [HI] in mol dm⁻³ (3 s.f.).
mol dm⁻³
Hint: rearrange — [HI]² = Kc × [H₂][I₂] = 54.0 × 0.0100 × 0.0100 = 0.00540. Now take the square root.
Quick check
A catalyst and Kc
8A catalyst is added to a reaction mixture already at equilibrium. What happens?
Industry
Industrial compromise: the Haber process
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ mol⁻¹typical conditions: 450 °C, 200 atm, iron catalyst
Temperature — a compromise. A low temperature would give a higher yield (the forward reaction is exothermic), but the rate would be far too slow and it would take a very long time to reach equilibrium. 450 °C sacrifices some yield for an acceptable rate.
Pressure — a compromise. A high pressure gives a higher yield (4 mol → 2 mol) and a faster rate, but very high pressures need thick-walled vessels and huge compression energy, which is expensive and hazardous. 200 atm balances yield against cost and safety.
Catalyst — no compromise needed. Iron increases the rate without affecting the yield, and it allows a lower temperature to be used than would otherwise be practical.
Recycling. The ammonia is condensed out (it liquefies well above the boiling points of N₂ and H₂) and the unreacted gases are recycled, so the overall conversion is high even though the yield per pass is only about 15%.
Industry
Industrial compromise: the Contact process
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJ mol⁻¹typical conditions: 450 °C, about 1–2 atm, V₂O₅ catalyst
Temperature — same compromise as the Haber process: exothermic forward reaction, so 450 °C trades a little yield for a workable rate.
Pressure — here is the interesting difference. There are 3 mol of gas on the left and 2 on the right, so high pressure would improve the yield — but the yield at just 1–2 atm is already about 96%. Raising the pressure would add enormous cost for a negligible gain, so it is not worth it.
Catalyst — vanadium(V) oxide, V₂O₅, a heterogeneous catalyst. It speeds up the reaction; it does not change the yield.
The examiner's point: a compromise is only made where a genuine conflict exists. For temperature there is always a conflict (yield vs rate) in an exothermic reaction. For pressure in the Contact process there is no conflict worth resolving, because the yield is already almost complete.
Quick check
Why 450 °C?
9The Haber process runs at about 450 °C even though a lower temperature would give a higher equilibrium yield of ammonia. Why?
Quick check
Why not high pressure in the Contact process?
10The Contact process uses only about 1–2 atm, even though 3 mol of gas become 2 mol. Why is high pressure not used?
Sort it
Which way does it shift?
The system is N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ mol⁻¹, at equilibrium in a closed vessel. Tap a change, then tap what it does to the position.
➡️ Shifts right (more NH₃)
⬅️ Shifts left (less NH₃)
⏸️ No shift
Match it
Position of equilibrium vs the value of Kc
The reaction is exothermic in the forward direction, with fewer moles of gas on the right. Tap an item on the left, then its partner on the right.