Part 2: Kc calculations and industrial choices

Write equilibrium expressions from balanced equations, use supplied equilibrium concentrations, and evaluate rate–yield compromises.

Kc is required in OCR Year 12

For a homogeneous equilibrium aA + bB ⇌ cC + dD, Kc = [C]ᶜ[D]ᵈ / ([A]ᵃ[B]ᵇ). Square brackets mean equilibrium concentration. Coefficients become powers, not multipliers outside the brackets. Use the given equilibrium concentrations; initial concentrations are not interchangeable.

For N₂ + 3H₂ ⇌ 2NH₃: Kc = [NH₃]²/([N₂][H₂]³). For H₂ + I₂ ⇌ 2HI: Kc = [HI]²/([H₂][I₂]). The AS outcome does not require deriving Kc units. More involved equilibrium-amount calculations and heterogeneous expressions are developed in Year 13.

Build the expression directly from the balanced equation

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), write product concentration squared on top, then sulfur dioxide concentration squared multiplied by oxygen concentration below: Kc = [SO₃]²/([SO₂]²[O₂]). Brackets mean concentration, not a number of molecules or the chemical formula’s relative mass.

A stoichiometric coefficient becomes a power. The subscript 2 inside SO₂ identifies sulfur dioxide and does not tell you to square its concentration; it is the separate coefficient 2 before SO₂ that supplies that power. A coefficient of one needs no written power.

Use concentrations already at equilibrium. If a question gives them in mol dm⁻³ and also gives vessel volume, do not divide by volume again. More involved initial-to-equilibrium amount tables and Kc units are Year 13 work; AS calculations here use the supplied equilibrium concentrations.

Worked Kc example

For H₂(g) + I₂(g) ⇌ 2HI(g), equilibrium concentrations are [H₂] = 0.200, [I₂] = 0.100 and [HI] = 0.800 mol dm⁻³. Kc = 0.800²/(0.200 × 0.100) = 32.0 to three significant figures. Do the powers before division and retain the full denominator.

A larger Kc indicates a more product-favoured equilibrium for that written reaction; a small value indicates reactant-favoured equilibrium. It says nothing about how quickly equilibrium is reached. Reversing the equation inverts Kc; compare values only with the reaction definition in mind.

At a fixed temperature, concentration or pressure changes alter equilibrium composition but not Kc. Temperature changes can alter Kc: raising temperature reduces Kc for an exothermic forward reaction and increases it for an endothermic one.

Calculate Kc when the powers do not cancel

Constructed example for 2SO₂ + O₂ ⇌ 2SO₃: equilibrium concentrations are SO₂ 0.200, O₂ 0.100 and SO₃ 0.400 mol dm⁻³. Numerator = 0.400² = 0.160. Denominator = 0.200² × 0.100 = 0.00400. Therefore Kc = 0.160/0.00400 = 40.0 to three significant figures.

Enter the whole denominator in brackets. Typing 0.400²/0.200² × 0.100 calculates a different expression because multiplication and division are processed in sequence. Writing the substituted expression before using the calculator exposes that mistake.

For A ⇌ B with [A] = 0.800 and [B] = 0.0200, Kc = 0.0250 = 2.50 × 10⁻²; the reactant is favoured. For this 1:1 example, Kc = 1 means equal concentrations. That conclusion is not general for equations with different powers: Kc = 1 means the specified concentration products have equal values.

Choose conditions for useful output

For an exothermic gas equilibrium such as ammonia synthesis, lower temperature gives a higher equilibrium yield but a slower reaction. A moderate temperature with a catalyst balances conversion and production rate. Higher pressure favours ammonia and raises collision frequency, but compression energy, strong equipment and safety costs limit how far it is worthwhile.

Removing product and recycling unreacted feed can improve overall conversion over repeated passes; this does not mean every single pass reaches 100% conversion. Use the equation and economic information provided rather than memorising one factory’s exact conditions.

Compare product per unit time, energy use, feedstock cost, separation, catalyst lifetime, waste and risk. A condition that maximises equilibrium yield alone need not produce the cheapest or most sustainable process.

Explain why maximum yield is not the sole design target

An industrial reactor must produce a useful amount per unit time, safely and at acceptable cost. For exothermic ammonia synthesis, very low temperature may improve equilibrium conversion but make the uncatalysed approach too slow. A catalyst increases rate and enables operation at a more useful temperature, while the chosen temperature still determines Kc.

Higher pressure favours ammonia because four gas moles become two, but compression costs energy and equipment must withstand the pressure. Product separation followed by recycling allows unreacted nitrogen and hydrogen to have another pass. Overall conversion across repeated passes can therefore exceed conversion in a single pass without claiming the equilibrium has disappeared.

For an unfamiliar industry question, first annotate forward ΔH and gas coefficients, then distinguish rate, equilibrium yield and engineering costs. Use supplied cost or yield data to choose conditions; do not transplant memorised temperatures or pressures from an unrelated process.

Quick checks

Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.

Sources