Reversible reactions

· Reversible reaction = a reaction where products can react to reform reactants.
· Shown using the equilibrium arrow: ⇌.
· Example: N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
· The forward reaction goes from reactants → products.
· The reverse reaction goes from products → reactants.
· A reversible reaction can reach dynamic equilibrium only in a closed system.

Dynamic equilibrium

· Dynamic equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction.
· At equilibrium, the concentrations of reactants and products remain constant.
· Constant concentration does not mean equal concentration.
· Reactions continue at equilibrium, so equilibrium is dynamic, not static.
· A closed system is needed so no reactants or products escape.

This graph shows reactant and product concentrations becoming constant over time. It is useful for remembering that equilibrium is reached when concentrations stop changing, not when they become equal. Source

Closed systems

· A closed system means no substances can enter or leave.
· Dynamic equilibrium cannot be established if gases or volatile substances escape.
· In exam answers, always link closed system to both forward and reverse reactions continuing.
· Example: a sealed container allows NO₂(g) and N₂O₄(g) to interconvert until equilibrium is reached.

Le Chatelier’s principle

· Le Chatelier’s principle: if a change is made to a system at dynamic equilibrium, the position of equilibrium moves to minimise this change.
· The equilibrium can shift to the right = more products formed.
· The equilibrium can shift to the left = more reactants formed.
· Exam wording: say “the position of equilibrium shifts…”, not “the reaction shifts”.
· Le Chatelier predictions are qualitative: they describe direction and effect, not exact quantities.

Effect of concentration changes

· Increasing concentration of a reactant shifts equilibrium to the right to use up the added reactant.
· Increasing concentration of a product shifts equilibrium to the left to use up the added product.
· Removing a reactant shifts equilibrium to the left to replace it.
· Removing a product shifts equilibrium to the right to replace it.
· The value of Kc or Kp is unchanged by concentration changes, provided temperature is constant.

Effect of pressure changes

· Pressure changes only affect equilibria involving gases.
· Increasing pressure shifts equilibrium to the side with fewer moles of gas.
· Decreasing pressure shifts equilibrium to the side with more moles of gas.
· If both sides have the same number of gaseous moles, pressure has no effect on equilibrium position.
· The value of Kp is unchanged by pressure changes, provided temperature is constant.

This diagram shows why increasing pressure favours the side with fewer gas particles. It is directly relevant to CIE questions asking for pressure effects in gaseous equilibria. Source

Effect of temperature changes

· Treat heat as if it is on one side of the equilibrium equation.
· For an exothermic forward reaction, heat is a product.
· Increasing temperature favours the endothermic direction.
· Decreasing temperature favours the exothermic direction.
· Temperature is the only listed factor here that changes the value of Kc or Kp.
· For an exothermic forward reaction, increasing temperature decreases K.
· For an endothermic forward reaction, increasing temperature increases K.

Effect of catalysts

· A catalyst has no effect on the position of equilibrium.
· A catalyst has no effect on Kc or Kp.
· A catalyst increases the rates of the forward and reverse reactions equally.
· A catalyst allows equilibrium to be reached faster.
· Common exam phrase: “same equilibrium yield, reached in less time.”

Equilibrium constants: Kc

· Kc = equilibrium constant in terms of concentration.
· Concentrations are written using square brackets: [ ].
· For aA + bB ⇌ cC + dD:
· Kc = [C]ᶜ[D]ᵈ / [A]ᵃ[B]ᵇ.
· Only include species whose concentrations change significantly in the equilibrium mixture.
· In CIE 7.1, calculations with Kc will not require solving quadratic equations.
· A larger Kc means equilibrium lies more towards products.
· A smaller Kc means equilibrium lies more towards reactants.

Mole fraction and partial pressure

· Mole fraction = moles of one gas ÷ total moles of gas.
· Mole fraction has no units.
· Partial pressure = pressure exerted by one gas in a mixture.
· Partial pressure = mole fraction × total pressure.
· The sum of all partial pressures equals the total pressure.

Equilibrium constants: Kp

· Kp = equilibrium constant in terms of partial pressures.
· For aA(g) + bB(g) ⇌ cC(g) + dD(g):
· Kp = (pC)ᶜ(pD)ᵈ / (pA)ᵃ(pB)ᵇ.
· Only include gaseous species in Kp expressions.
· Do not use the relationship between Kp and Kc for this syllabus section.
· In calculations, first find moles at equilibrium, then mole fractions, then partial pressures, then substitute into Kp.

Calculating equilibrium quantities

· Start with a balanced equation and set up an initial-change-equilibrium table.
· Use the stoichiometric ratio to calculate how each amount changes.
· Convert equilibrium amounts into concentrations for Kc questions.
· Convert equilibrium amounts into mole fractions and partial pressures for Kp questions.
· Always check whether the question asks for amount, concentration, partial pressure, Kc or Kp.
· Use units only when appropriate; mole fraction has no units.

What affects the value of K?

· Temperature changes Kc and Kp.
· Concentration does not change Kc or Kp.
· Pressure does not change Kc or Kp.
· Catalysts do not change Kc or Kp.
· If temperature is constant, the system may shift position, but K remains constant.

Haber process

· Reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g).
· Forward reaction is exothermic.
· High pressure favours ammonia because the product side has fewer moles of gas: 4 mol gas → 2 mol gas.
· Lower temperature favours ammonia yield because the forward reaction is exothermic.
· A very low temperature is too slow, so industry uses a compromise temperature.
· Iron catalyst increases rate but does not increase equilibrium yield.
· Unreacted N₂ and H₂ are recycled to improve overall yield.
· Key exam idea: industrial conditions balance yield, rate and cost.

This flow scheme shows how nitrogen and hydrogen are converted into ammonia, with unreacted gases recycled. It links dynamic equilibrium to industrial optimisation of yield, rate and cost. Source

Contact process

· Key equilibrium step: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g).
· Forward reaction is exothermic.
· Higher pressure favours SO₃ because the product side has fewer moles of gas: 3 mol gas → 2 mol gas.
· A very high pressure is usually not needed because yield is already high at moderate pressure.
· Lower temperature favours SO₃ yield, but too low a temperature gives a slow rate.
· V₂O₅ catalyst increases rate but does not change equilibrium position or Kp.
· Key exam idea: conditions are chosen as a compromise between yield, rate and economics.

This page supports the industrial equilibrium example for sulfur trioxide manufacture. It shows how Le Chatelier’s principle explains the choice of pressure, temperature and catalyst in the Contact process. Source

Common exam traps

· Do not say equilibrium means equal amounts; it means equal rates.
· Do not say a catalyst increases yield; it only increases rate of reaching equilibrium.
· Do not include solids or liquids in Kp expressions; Kp uses gaseous partial pressures.
· Do not say pressure affects all equilibria; it only matters when there are gases and different gaseous mole numbers.
· Do not say concentration or pressure changes K; only temperature changes K.