Chemical Equilibria: Kc, Kp and Le Chatelier's Principle

This guide covers subtopic 7.1, Chemical equilibria: reversible reactions, dynamic equilibrium, from Topic 7, Equilibria, of Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. This is AS Level content.

Before studying this

At IGCSE or O Level, you met reversible reactions and the qualitative statement of Le Chatelier’s principle, applied to the Haber and Contact processes. Rates of Reaction and Reversible Reactions covers that treatment, including the actual conditions used industrially.

AS Level keeps the same two industrial examples but adds the quantitative side entirely: writing and using equilibrium-constant expressions (Kc for concentrations, Kp for partial pressures), and using them — not just Le Chatelier’s principle in words — to reason about what does and does not shift an equilibrium constant’s value. This page assumes the qualitative picture from the IGCSE resource is already familiar.

Syllabus coverage

CAMBRIDGE INTERNATIONAL AS & A LEVEL CHEMISTRY 9701 — AS Level, Topic 7

7.1 Chemical equilibria: reversible reactions, dynamic equilibrium — understanding what is meant by a reversible reaction and by dynamic equilibrium (rate of forward and reverse reactions equal, concentrations of reactants and products constant), and the need for a closed system; defining Le Chatelier’s principle; using Le Chatelier’s principle to deduce qualitatively the effects of changes in temperature, concentration, pressure or catalyst on a system at equilibrium; deducing expressions for equilibrium constants in terms of concentration, Kc; using the terms mole fraction and partial pressure; deducing expressions for equilibrium constants in terms of partial pressure, Kp; using Kc and Kp expressions in calculations (not requiring quadratic equations); calculating the quantities present at equilibrium given appropriate data; stating whether changes in temperature, concentration, pressure or catalyst affect the value of the equilibrium constant; describing and explaining the conditions used in the Haber process and the Contact process as examples of dynamic equilibrium and Le Chatelier’s principle in the chemical industry.

Dynamic equilibrium

A reversible reaction can proceed in both the forward and reverse directions. In a closed system — nothing entering or leaving — a reversible reaction reaches dynamic equilibrium: the forward and reverse reactions continue happening, but at exactly equal rates, so the concentrations of reactants and products stop changing. It’s dynamic, not static — the reaction hasn’t stopped, the two directions have simply balanced.

Le Chatelier’s principle: if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise that change.

Applying Le Chatelier’s principle

• Concentration: increasing a reactant’s concentration shifts the position of equilibrium towards the products (using up some of the added reactant); increasing a product’s concentration shifts it back towards the reactants.
• Pressure (gas equilibria only): increasing pressure shifts the position of equilibrium towards the side with fewer moles of gas, since that reduces the total number of gas particles and opposes the pressure increase.
• Temperature: increasing temperature shifts the position of equilibrium in the endothermic direction, since that absorbs some of the added heat.
• Catalyst: a catalyst speeds up the forward and reverse reactions equally, so it has no effect on the position of equilibrium — it only gets the system to equilibrium faster.

Kc: the equilibrium constant in terms of concentration

For a general reaction aA + bB ⇌ cC + dD at equilibrium:

Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ

with concentrations in mol dm⁻³ and each raised to the power of its balancing number.

Worked example. At equilibrium in a 1 dm³ sealed container, the reaction H₂(g) + I₂(g) ⇌ 2HI(g) contains 0.50 mol H₂, 0.50 mol I₂ and 3.0 mol HI. Calculate Kc.

Since the volume is 1 dm³, concentrations equal the mole values directly.

Kc = [HI]² / ([H₂][I₂]) = (3.0)² / (0.50 × 0.50) = 9.0 / 0.25 = 36

There are equal total moles of gas on each side (1+1 → 2), so the concentration units cancel exactly and Kc has no units.

Mole fraction, partial pressure and Kp

For a mixture of gases, the mole fraction of component A is the proportion of the total moles it represents:

xₐ = nₐ / n(total)

Its partial pressure is that mole fraction multiplied by the total pressure:

Pₐ = xₐ × P(total)

Kp is written the same way as Kc, but using partial pressures in place of concentrations:

Kp = (P꜀)ᶜ(P_D)ᵈ / (Pₐ)ᵃ(P_B)ᵇ

Worked example. A sealed vessel at total pressure 200 kPa contains an equilibrium mixture of N₂O₄(g) ⇌ 2NO₂(g) with mole fractions 0.60 for N₂O₄ and 0.40 for NO₂. Calculate Kp.

Partial pressures: P(N₂O₄) = 0.60 × 200 = 120 kPa; P(NO₂) = 0.40 × 200 = 80 kPa.

Kp = (P_NO₂)² / P_N₂O₄ = (80)² / 120 = 6400 / 120 = 53.3 kPa

Here the mole count changes (1 → 2), so the pressure units don’t cancel — Kp carries units, here kPa¹, because the net order is (2 − 1) = 1.

What does and doesn’t change K

This is a common source of confusion, because Le Chatelier’s principle describes position of equilibrium shifting, while a completely separate question is whether the equilibrium constant itself changes:

• Changing concentration, pressure, or adding a catalyst shifts the position of equilibrium (or, for a catalyst, nothing at all) but leaves the value of K unchanged, provided temperature is constant.
• Only changing temperature changes the value of K.

The Haber and Contact processes

Both are textbook applications of balancing equilibrium yield against reaction rate and cost — a genuine test of applying Le Chatelier’s principle to an industrial decision, not just stating it.

Haber process: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH negative (exothermic). Fewer moles of gas on the product side favours high pressure for yield, and the exothermic forward reaction favours low temperature for yield — but low temperature makes the reaction too slow to be economic. The industrial compromise is roughly 450 °C and 200 atm, with an iron catalyst to increase the rate at which equilibrium is reached (the catalyst does not affect the equilibrium yield itself).

Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), also exothermic. The same logic applies, but because the equilibrium yield is already very high (around 99.5%) at close to atmospheric pressure, the extra yield from higher pressure isn’t worth its extra cost — so the Contact process runs at close to atmospheric pressure, around 450 °C, with a vanadium(V) oxide catalyst.

Common mistakes

• Treating “no effect on K” and “no effect on yield” as the same statement. A catalyst has no effect on either. But concentration and pressure changes do change the position of equilibrium (and therefore the yield) — they just don’t change the numerical value of K.
• Forgetting to raise concentrations/pressures to the power of their balancing number in Kc/Kp. Every term in the expression is raised to the power of its coefficient in the balanced equation, not left to the first power by default.
• Assuming Kc always has no units. It only has no units when the total moles of gas (or dissolved species) are equal on both sides — check the balanced equation before assuming.
• Explaining industrial conditions using yield alone. Both the Haber and Contact processes are chosen as a compromise between equilibrium yield and a practical reaction rate/cost — a full answer addresses both.

Quick revision checklist

• Dynamic equilibrium: equal forward/reverse rates, constant concentrations, closed system required
• Le Chatelier’s principle and its application to concentration, pressure, temperature and catalysts
• Writing and calculating Kc and Kp expressions, including units
• Mole fraction and partial pressure
• What changes K (temperature only) vs what changes position only (concentration, pressure, catalyst)
• Haber and Contact process conditions as a yield/rate compromise

Related resources

• Rates of Reaction and Reversible Reactions — the IGCSE/O Level qualitative treatment this resource builds on
• Acids and Bases: The Brønsted-Lowry Theory — the other AS Level equilibria topic
• Reaction Kinetics: Collision Theory and Catalysis — why a catalyst speeds up the approach to equilibrium without changing it
• Redox Processes: Oxidation Numbers and Electron Transfer — the Contact process is itself a redox reaction
• Acids, Bases, Buffers and Partition Coefficients — the A Level topic that extends equilibrium constants to Ka, Ksp and Kpc
• Cambridge AS & A Level Chemistry hub

Written against Cambridge International AS & A Level Chemistry 9701, 2025–2027 series. Always check the current syllabus for your examination year.