Part 1: Equilibrium amounts, Kc and experimental evidence

Build an equilibrium calculation from the balanced equation and the quantities actually present. Then use a titration to infer an equilibrium composition.

Start with a changing reaction and a constant composition

A closed reacting mixture reaches dynamic equilibrium when the forward and reverse reactions occur at equal rates. Molecules continue reacting, but each substance is produced as quickly as it is consumed, so its concentration stays constant. The two rates are equal; the concentrations usually are not.

The balanced equation describes the proportions of amounts consumed and formed, not a compulsory ratio of final amounts. Before finding Kc, distinguish the initial amount, the change in amount and the equilibrium amount. Every dissolved substance or gas occupies the whole relevant solution or vessel volume.

For aA + bB ⇌ cC + dD, the concentration expression is Kc = [C]ᶜ[D]ᵈ/([A]ᵃ[B]ᵇ). Insert equilibrium concentrations in mol dm⁻³. The expression refers to the equation as written and a stated temperature. Reversing an equation gives 1/K; multiplying every coefficient by two gives K².

Worked example: product is already present

In a 4.00 dm³ vessel, H₂(g) + I₂(g) ⇌ 2HI(g) starts with 0.700 mol H₂, 0.500 mol I₂ and 0.200 mol HI. Equilibrium analysis gives 0.800 mol HI. Its increase is 0.600 mol, so 0.300 mol of each reactant has reacted.

The final concentrations are 0.400/4.00 = 0.100 mol dm⁻³ H₂, 0.200/4.00 = 0.0500 mol dm⁻³ I₂ and 0.800/4.00 = 0.200 mol dm⁻³ HI. Thus Kc = 0.200²/(0.100 × 0.0500) = 8.00. Concentration units cancel: this expression has no units.

For a different equation, N₂O₄ ⇌ 2NO₂, Kc has units (mol dm⁻³)²/(mol dm⁻³) = mol dm⁻³. If [N₂O₄] = 0.160 and [NO₂] = 0.240 mol dm⁻³, Kc = 0.360 mol dm⁻³. A volume cannot be cancelled unless its total powers on the two sides of the expression are equal.

Homogeneous and heterogeneous systems

A homogeneous equilibrium contains one phase. A heterogeneous equilibrium contains more than one phase. For the concentration model used at this level, omit a separate pure solid or pure liquid phase: its effective activity is constant while that phase remains present. Include aqueous solutes whose concentrations can change.

For CaCO₃(s) ⇌ CaO(s) + CO₂(g), Kc = [CO₂]. Adding extra carbonate does not change the equilibrium CO₂ concentration at fixed temperature when both solids remain. If one solid has been completely consumed, the assumed two-solid equilibrium no longer applies.

Water as the solvent in a dilute aqueous equilibrium is omitted. Water in a homogeneous esterification mixture may be an appreciably varying component and is included in the supplied Kc model. Distinguish the physical situation rather than applying “omit every liquid” to all mixtures.

From a titration to an equilibrium composition

An esterification investigation mixes known quantities of an alcohol and carboxylic acid with an acid catalyst. Use stoppered vessels to limit volatile loss, maintain a chosen temperature and leave replicate mixtures for increasing times. Agreement between compositions at later times supports the conclusion that equilibrium has been reached; concordant titres of one early sample do not establish it.

An aliquot is analysed with standard sodium hydroxide. The titre measures both remaining carboxylic acid and the acid catalyst, so determine a catalyst blank with the same catalyst amount and dilution. Correct the titre before converting the result into an amount of carboxylic acid. Sulfuric acid contributes two acid equivalents per mole when fully neutralised.

Use a validated rapid cooling/dilution and analysis procedure to make further esterification or hydrolysis negligible during measurement. Dilution disturbs equilibrium, so it is not justified to claim that the diluted solution has the original equilibrium composition indefinitely. Rinse transfer vessels quantitatively, use a volumetric pipette, and repeat titres to a defined concordance criterion.

Original data: 10.00 cm³ of an analysed mixture requires 18.40 cm³ of 0.1000 mol dm⁻³ NaOH; a matching catalyst blank requires 2.40 cm³. Corrected acid amount = 0.01600 × 0.1000 = 0.001600 mol, so acid concentration = 0.001600/0.01000 = 0.1600 mol dm⁻³. If an original aliquot was diluted first, apply that stated dilution factor.

If the original equilibrium vessel contained 0.500 mol acid, 0.600 mol alcohol, 0 mol ester and 0.100 mol water initially, and analysis finds 0.200 mol acid remaining, 0.300 mol has reacted. Equilibrium amounts are acid 0.200, alcohol 0.300, ester 0.300 and water 0.400 mol. In the homogeneous-mixture model, Kc = (0.300 × 0.400)/(0.200 × 0.300) = 2.00 because common volume factors cancel.

Ignoring the acid catalyst overestimates remaining carboxylic acid; under the stated starting-mixture model this underestimates conversion and Kc. Volatile loss changes material balances and cannot be fixed by repeating titres. Keep flammable alcohol away from ignition sources and use suitable eye protection for acid and alkali. This is a possible investigation for Topic 11, not an additional numbered core practical.

Quick checks

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

Sources