Connect a real measurement to equilibrium amounts and explain concentration, temperature and catalyst changes without treating K as a moving position.
K constrains the equilibrium ratio
K fixes a particular combination of equilibrium concentrations or pressures at one temperature. A large value generally favours products relative to the balanced expression, but does not prove every reactant has disappeared. It says nothing directly about the time taken to reach equilibrium.
After adding a reactant at fixed temperature, the mixture initially no longer satisfies K. Net forward reaction changes the composition until the expression again equals the same K. For an original illustrative A ⇌ B equilibrium with Kc = [B]/[A] = 4, an increase in [A] initially lowers the ratio; net A → B restores it. The equilibrium concentrations change while the constant stays fixed.
This reaction-quotient reasoning is a useful explanatory method rather than an extra required symbol to memorise. Apply it consistently to Kc, Kp, Ka or another supplied equilibrium expression.
Temperature changes the constant itself
If the forward reaction is exothermic, raising temperature favours the reverse, endothermic direction and decreases the forward equilibrium constant. If the forward reaction is endothermic, raising temperature increases the constant. State the direction of the equation because reversing the equation reverses which process is exothermic.
A catalyst does not change the energies of reactants or products and therefore does not change the equilibrium constant. Concentration and compression at fixed temperature disturb the ratio and may change composition, but do not change K. Do not write “K moves right”: composition shifts, while a numerical K increases or decreases only when the relevant conditions alter it.
| Change | Equilibrium composition | Equilibrium constant |
|---|---|---|
| Add reactant at fixed T | Usually adjusts to consume some addition | Unchanged |
| Compress gases at fixed T | Depends on gas coefficients | Unchanged |
| Add catalyst | No change to final composition | Unchanged |
| Change temperature | Depends on reaction enthalpy | Changes |
Measure equilibrium composition without forgetting the catalyst
A liquid esterification equilibrium can be studied by mixing known amounts, maintaining a fixed temperature and allowing sufficient time to equilibrate. Analyse known samples until successive results are consistent; one early sample is not evidence that equilibrium has been reached. A sealed vessel limits loss of volatile material.
Titration of an equilibrium aliquot with standard alkali measures total titratable acid. If sulfuric acid was used as a catalyst, its contribution must be determined by a suitable blank or calculated from a known amount. Subtract those acid equivalents before assigning all the titre to the carboxylic acid. The aliquot-to-total-volume factor comes afterwards.
Original illustrative data: a 10.0 cm³ aliquot requires 14.60 cm³ of 0.1000 mol dm⁻³ NaOH. A catalyst blank for the same aliquot size requires 2.00 cm³. NaOH attributable to the monobasic organic acid is (14.60 − 2.00) × 0.1000/1000 = 1.260 × 10⁻³ mol. Its equilibrium concentration is 0.1260 mol dm⁻³.
Use the resulting loss of acid and balanced reaction ratio to find ester, alcohol and water amounts. Include water already present, where the chosen homogeneous-mixture model includes water in Kc. Account for sampling volume and state the model rather than silently treating the mixture as ideal in every circumstance.
Explain the direction of a specific error
Removing an aliquot and titrating it can allow further reaction during analysis. A suitable validated quench, such as rapid cooling and dilution for the chosen system, should make subsequent reaction negligible over the measurement time. Explain why the chosen quench is effective; cooling is not a universal chemical stop switch.
If the catalyst contribution is ignored, the inferred amount of remaining organic acid is too high. In a simple initially acid-and-alcohol mixture this makes inferred ester formation too low and can make calculated Kc too small. State the particular model before claiming a direction, because pre-existing products or a different analytical method may change the chain of reasoning.
Use a volumetric pipette for the aliquot, repeat concordant titrations and separate measurement uncertainty from systematic biases such as volatile loss. A burette titre is a difference of two readings, so both reading uncertainties contribute. Practical judgement is part of the chemistry, not an optional afterthought.
Quick checks
Original Finesse questions. Reveal the indicative worked solutions after attempting each question; these are not official OCR A mark allocations.
Q1. For an exothermic forward reaction, what happens to K on cooling?Show answer
Cooling favours the exothermic forward direction, so the product-to-reactant equilibrium expression increases. The forward K increases. State which written direction is exothermic.
Q2. An equilibrium has Kc = 5.0. Adding product at fixed temperature makes the current ratio 8.0. What follows?Show answer
Net reverse reaction consumes some product and forms reactant until the ratio returns to 5.0. Kc itself remains 5.0 at the unchanged temperature.
Q3. A 20.0 cm³ aliquot needs 22.50 cm³ of 0.0800 mol dm⁻³ NaOH; its catalyst blank needs 1.50 cm³. Find the monobasic organic acid concentration.Show answer
Corrected titre = 21.00 cm³. Acid amount = 0.02100 × 0.0800 = 0.001680 mol. Divide by 0.0200 dm³ to obtain 0.0840 mol dm⁻³.
Q4. Why is “the catalyst gives more product” misleading?Show answer
At a fixed short sampling time a catalyst may give more product because equilibrium is approached sooner. At equilibrium under the same conditions, the product proportion is unchanged. Distinguish rate from final yield.
Q5. A titration uses a single equilibrium sample taken soon after mixing. Give two improvements with reasons.Show answer
Allow sufficient time at controlled temperature and compare samples at later times to establish stable composition. Use an appropriate quench and analyse promptly so the sample composition does not appreciably change during titration. Replicate titres improve precision but cannot by themselves prove equilibrium.
Sources
Sources and examiner guidance (reviewed 6 October 2026)
- OCR A H432 specification — version 3.1 — 5.1.2, printed pp. 44; outcomes and additional guidance, with relevant Module 1 practical skills.
- Chemrevise — OCR A 5.1.2 — Pages 1–6; secondary coverage cross-check. Lesson explanations, data and questions are original Finesse material.
- OCR H432/01 mark scheme — June 2025 — Q17(c); printed pp. 17–18. Question-specific evidence, not universal marking rules.
- OCR H432/01 examiner report — June 2025 — Q17(c); printed pp. 30–32. Read with the corresponding question context.
- OCR H432/01 question paper — June 2025 — Q17(c); context for the assessment references, not reproduced questions.
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