Cambridge AS Level Chemistry 9701

Equilibria

A reaction is at dynamic equilibrium when the forward and reverse reactions run at the same rate, so the concentrations stop changing. Le Chatelier’s principle says that if conditions change, the position of equilibrium shifts to oppose the change. Only a change in temperature changes the equilibrium constant $K_{c}$ ; concentration and pressure move the position but leave $K_{c}$ alone, and a catalyst changes neither.

Cambridge AS Chemistry 9701 Paper 1 tests this mostly as “which change shifts the position, and which changes $K_{c}$ ” — plus Brønsted–Lowry acids and bases.

Updated 28 September 2026

How much it is worth

Equilibria: 49 questions across 23 real Cambridge 9701 papers — 5.3% of the total, and it came up on every one of them. How that ranks against every other 9701 topic →

Dynamic equilibrium

In a closed system a reversible reaction reaches a state where the forward and reverse reactions continue at equal rates. Both are still happening — that is what “dynamic” means — but the concentrations of reactants and products stay constant. Equal rates, not equal concentrations.

Le Chatelier's principle

Change	Position shifts	Kc
Add a reactant (or remove a product)	To the right	Unchanged
Increase pressure (gases)	Towards fewer gas molecules	Unchanged
Increase temperature	In the endothermic direction	Changes
Add a catalyst	No shift — equilibrium is reached faster	Unchanged

For an exothermic forward reaction, raising the temperature shifts the position to the left and decreases $K_{c}$ . For an endothermic forward reaction it does the opposite. The sign of $Δ H$ is given with the equation for exactly this reason — chemical energetics and equilibria meet here.

Industrial conditions are compromises. In the Haber process a lower temperature would give a higher equilibrium yield of ammonia, but the rate would be too slow, so a moderate temperature and an iron catalyst are used.

Kc and what changes it

For $a A + b B ⇌ c C + d D$ :

K_{c} = \frac{[C]^{c} [D]^{d}}{[A]^{a} [B]^{b}}

Products over reactants, each concentration (in mol dm⁻³) raised to its coefficient. Pure solids are left out. The units depend on the equation — work them out by cancelling, and when the powers top and bottom match, $K_{c}$ has no units.

A large $K_{c}$ means the equilibrium lies to the right. Only temperature changes its value.

Kp and partial pressures

For gases, the same expression is written with partial pressures. The partial pressure of a gas is its mole fraction times the total pressure:

p_{A} = \frac{n_{A}}{n_{total}} \times P K_{p} = \frac{p_{C}^{c} p_{D}^{d}}{p_{A}^{a} p_{B}^{b}}

Brønsted–Lowry acids and bases

A Brønsted–Lowry acid is a proton (H⁺) donor; a base is a proton acceptor. When an acid donates a proton, what is left is its conjugate base:

H C l + H_{2} O \to C l^{-} + H_{3} O^{+}

HCl and Cl⁻ are a conjugate pair; so are H₂O and H₃O⁺. Water is acting as a base here — with ammonia it acts as an acid, donating a proton to form NH₄⁺ and OH⁻.

A strong acid dissociates fully in water (HCl, HNO₃, H₂SO₄); a weak acid only partly (ethanoic acid), so its dissociation is itself an equilibrium. Strong and weak are about the extent of dissociation, not concentration.

Worked example

1.00 mol of hydrogen and 1.00 mol of iodine are sealed in a 1.00 dm³ flask and heated: H₂(g) + I₂(g) ⇌ 2HI(g). At equilibrium there are 1.56 mol of HI. Calculate Kc.

Forming 1.56 mol of HI uses 0.78 mol each of H₂ and I₂, leaving 0.22 mol of each.

K_{c} = \frac{[H I]^{2}}{[H_{2}] [I_{2}]} = \frac{{1.56}^{2}}{0.22 \times 0.22} = \frac{2.4336}{0.0484} = 50.3

No units: two moles on each side, so they cancel. For the same reason the volume cancels too — which is why a question can leave it out for this equation but not for one where the moles differ.

Common mistakes

1.Saying equilibrium means equal concentrations.
It means equal rates. The concentrations are constant, not equal.
2.Saying a pressure change alters Kc.
It shifts the position; $K_{c}$ changes only with temperature.
3.Saying a catalyst increases the yield.
A catalyst speeds up the forward and reverse reactions equally. Equilibrium arrives sooner; the position is unchanged.
4.Using moles instead of concentrations in Kc.
Divide by the volume first — unless, as in the example, the moles on both sides are equal and it cancels.
5.Confusing strong with concentrated.
A dilute solution of a strong acid is still fully dissociated. Strength is extent of dissociation; concentration is amount per volume.

Common questions

What changes the value of Kc?

Only temperature. Changing concentration or pressure moves the position of equilibrium but $K_{c}$ stays the same; a catalyst changes neither.

Why does increasing temperature lower the yield of an exothermic reaction?

The system opposes the rise by favouring the endothermic direction, which for an exothermic forward reaction is the reverse reaction — so less product at equilibrium, and a smaller $K_{c}$ .

What is a conjugate acid–base pair?

Two species that differ by one proton, such as NH₄⁺ and NH₃, or H₂O and OH⁻.

Equilibria on real papers

The papers that leaned on it hardest, as a share of the paper. Each one has its own page on Quanta with the full topic breakdown:

9701/11/M/J/253 of 40 questions 9701/14/M/J/253 of 40 questions 9701/12/M/J/243 of 40 questions

Practise equilibria against real mark schemes

Quanta has real Cambridge AS Chemistry 9701 past-paper questions, with every answer explained after you commit to it and each question tied to the syllabus topic it tests — and it tracks which skills you’re missing. Free for individual students.

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