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Real Past Paper Questions — With Full Memos
Attempt each question before revealing the memo. Questions drawn from Heron Bridge, Hilton, St John's, DSG, Waverley & IEB 2021.
Heron Bridge 2021
Q5 — Titration of Ba(OH)₂ with (COOH)₂ · Diprotic Acids · Hydrolysis
Ba(OH)₂ is a strong alkali. (COOH)₂ (oxalic acid) is a weak diprotic acid. 25 cm³ of alkali is placed in flask Y. Phenolphthalein indicator is used. 18,6 cm³ of acid (0,2 mol·dm⁻³) is needed to neutralise 25 cm³ of the alkali.
5.1 (2) What is meant by saying Ba(OH)₂ is a strong alkali?
5.2 (1) What is a diprotic acid?
5.3 (1) Name the dilute alkali.
5.4 (1) Name the apparatus labelled X (the burette in the diagram).
5.5 (2) What is meant by the equivalence point of the titration?
5.6.1 (2) Write a balanced chemical equation for the neutralisation reaction between Ba(OH)₂ and (COOH)₂.
5.6.2 (3) Calculate the number of moles of alkali available at the start of the reaction.
5.6.3 (3) Hence calculate the concentration of the acid.
5.7.1 (1) State whether [Ba²⁺] INCREASES, DECREASES or REMAINS CONSTANT while acid is added before the end point.
5.7.2 (1) State whether [OH⁻] INCREASES, DECREASES or REMAINS CONSTANT.
5.7.3 (1) State whether pH INCREASES, DECREASES or REMAINS CONSTANT.
5.8.1 (4) Complete the hydrolysis equations: NH₄⁺(aq) + H₂O(ℓ) → and CH₃COO⁻(aq) + H₂O(ℓ) →
5.8.2 (1) Name the phenomenon described in 5.8.1.
✓ Full Memo — Heron Bridge 2021
5.1 Strong alkali — dissociates completely in an aqueous solution. [2: 'completely' + 'aqueous solution']
5.2 Diprotic acid — able to donate two protons (H⁺) per molecule of acid. [1]
5.3 Barium hydroxide. [1]
5.4 Burette. [1]
5.5 The point where an acid and base (alkali) have reacted so that neither is in excess. [2]
5.6.1 Ba(OH)₂ + (COOH)₂ → (COO)₂Ba + 2H₂O [2: correct formulas + balanced]
5.6.2 c = n/V → n = cV = 0,2 × 0,025 = 5 × 10⁻³ mol [3: formula + sub + answer]
5.6.3 Ratio n(alkali) : n(acid) = 1:1 (from equation) ∴ n(acid) = 5×10⁻³ mol. c = n/V = (5×10⁻³)/0,0186 = 0,27 mol·dm⁻³ [3]
5.7.1 REMAINS CONSTANT — Ba²⁺ is a spectator ion and is not consumed. [1]
5.7.2 DECREASES — OH⁻ is being neutralised by the acid. [1]
5.7.3 DECREASES — as OH⁻ decreases, [H₃O⁺] increases, so pH falls. [1]
5.8.1 NH₄⁺(aq) + H₂O(ℓ) → NH₄OH + H⁺ and CH₃COO⁻(aq) + H₂O(ℓ) → CH₃COOH + OH⁻ [4: 1 mark each product, both equations]
5.8.2 Hydrolysis of a salt. [1]
Hilton College 2021
Q5 — Strong/Weak Acids · H₃PO₄ Polyprotic · Kw Calculation · Conductivity
Two circuits use 1 mol·dm⁻³ HCl(aq) and 1 mol·dm⁻³ CH₃CO₂H(aq) as electrolytes. Bulb 1 (HCl) is brighter. H₃PO₄ has concentration 1,67 × 10⁻³ mol·dm⁻³ in a soft drink. 3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O. 30 cm³ of 2 mol·dm⁻³ NaOH is titrated; 15 cm³ of acid was used.
5.1.1 (2) Define a strong acid.
5.1.2 (3) Bulb 1 (HCl) was brighter than Bulb 2 (CH₃CO₂H). Fully explain why.
5.1.3(i) (1) Why is H₃PO₄ considered polyprotic?
5.1.3(ii) (4) Write the balanced ionisation reaction of H₃PO₄ assuming complete ionisation.
5.1.3(iii) (2) Calculate [H₃O⁺] in the soft drink. (Assume complete ionisation.)
5.1.3(iv) (3) Hence calculate [OH⁻] at 25°C.
5.2.1 (2) Define standard solution.
5.2.2 (5) Calculate the concentration of H₃PO₄ used in the NaOH titration.
5.2.3 (1) Name the salt formed when NaOH reacts with H₃PO₄.
5.2.4 (2) Define hydrolysis of a salt.
5.2.5 (1) From the indicator table (Methyl red 3.1–4.4; Bromothymol blue 6.0–7.6; Phenolphthalein 8.4–11) name a suitable indicator for this titration.
5.2.6 (4) Fully justify your choice of indicator by considering hydrolysis of the salt. Use balanced equations.
✓ Full Memo — Hilton College 2021
5.1.1 An acid that ionises completely in solution. [2: 'ionises' + 'completely']
5.1.2 Bulb 1 has HCl — strong acid, ionises completely → many free ions → greater current → brighter bulb. Bulb 2 has CH₃CO₂H — weak acid, only partially ionises → very few ions → less current → dimmer bulb. [3: strong/weak id + ion comparison + brightness conclusion]
5.1.3(i) H₃PO₄ can donate more than one proton. [1]
5.1.3(ii) H₃PO₄ + 3H₂O ⇌ 3H₃O⁺ + PO₄³⁻ [4: correct formulas + balanced + reactants]
5.1.3(iii) Ratio H₃PO₄ : H₃O⁺ = 1:3. [H₃O⁺] = 3 × 1,67×10⁻³ = 5,01 × 10⁻³ mol·dm⁻³ [2]
5.1.3(iv) Kw = [H₃O⁺][OH⁻] = 1×10⁻¹⁴. [OH⁻] = 1×10⁻¹⁴ ÷ 5,01×10⁻³ = 2×10⁻¹² mol·dm⁻³ [3: Kw formula + sub + answer]
5.2.1 A solution of known concentration. [2]
5.2.2 n(NaOH) = 2 × 30/1000 = 0,06 mol. Ratio NaOH:H₃PO₄ = 3:1. n(H₃PO₄) = 0,06÷3 = 0,02 mol. c = n/V = 0,02 ÷ 0,015 = 1,33 mol·dm⁻³ [5]
5.2.3 Sodium phosphate. [1]
5.2.4 The reaction of an ion (from a salt) with water. [2]
5.2.5 Phenolphthalein. [1]
5.2.6 Salt = Na₃PO₄ (strong base + weak acid). PO₄³⁻ is the conjugate base of weak acid H₃PO₄. It hydrolyses: PO₄³⁻ + H₂O ⇌ OH⁻ + H₃PO₄. Excess OH⁻ → solution is basic → pH > 7 (8–10). Phenolphthalein (pH range 8.4–11) covers this range. [4]
St John's College 2021
Q6 — KOH vs H₂SO₄ · [OH⁻] from Kw · Volume Titration · Acid/Base Mix
0,025 mol·dm⁻³ KOH is titrated with 25 ml of 0,010 mol·dm⁻³ H₂SO₄. Equation: 2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(ℓ)
6.1.1 (2) Provide the Brønsted-Lowry definition for a base.
6.1.2 (2) Sulfuric acid is a strong acid. What does this mean?
6.1.3 (4) Calculate [OH⁻] in the original H₂SO₄ solution before addition of KOH.
6.1.4 (2) Define equivalence point.
6.1.5 (3) Determine the volume of KOH required to fully neutralise the H₂SO₄.
6.2 (2) After the endpoint, a few more drops of acid fall in. Explain how this affects the calculated concentration of the base.
6.3 (5) 2,5 ml of 0,525 mol·dm⁻³ NaOH and 7,5 ml of 0,355 mol·dm⁻³ HCl are accidentally mixed. Is the resulting solution acidic or basic? Show calculations.
✓ Full Memo — St John's College 2021
6.1.1 A base is a proton acceptor. [2: all or nothing]
6.1.2 The acid will completely ionise in solution. [2]
6.1.3 H₂SO₄ → 2H⁺, so [H₃O⁺] = 2 × 0,010 = 0,020 mol·dm⁻³. Kw = [H₃O⁺][OH⁻]. [OH⁻] = 1×10⁻¹⁴ ÷ 0,020 = 5,00×10⁻¹³ mol·dm⁻³ [4]
6.1.4 The point where acid and base have reacted so neither is in excess. [2]
6.1.5 n(H₂SO₄) = 25×10⁻³ × 0,010 = 2,50×10⁻⁴ mol. n(KOH) = 2 × 2,50×10⁻⁴ = 5,00×10⁻⁴ mol. V = n/c = 5,00×10⁻⁴ ÷ 0,025 = 0,020 dm³ = 20 cm³ [3]
6.2 The calculated concentration of the base will be higher than the actual concentration — extra acid means more base appears to have reacted. [2]
6.3 NaOH + HCl → NaCl + H₂O. n(NaOH) = 0,525 × 2,5×10⁻³ = 1,31×10⁻³ mol. n(HCl) = 0,355 × 7,5×10⁻³ = 2,66×10⁻³ mol. HCl is in excess → [H₃O⁺] > [OH⁻] → solution is acidic. [5]
DSG 2021
Q4 — NaOH Standard Solution · Dilute vs Strong · Back Titration · CaCl₂ Hydrolysis
2 g of pure NaOH is dissolved in a 250 cm³ volumetric flask. 1,5 g of CaCO₃ reacts with 50 cm³ of dilute HCl. The excess HCl is neutralised with 25 cm³ of the NaOH solution. Reaction A: 2HCl + CaCO₃ → CaCl₂ + CO₂ + H₂O. Reaction B: HCl + NaOH → NaCl + H₂O.
4.1 (1) Write the term for "a solution of known concentration".
4.2 (3) Write the dissociation equation for NaOH in water (include phase symbols).
4.3 (4) Calculate the concentration of the NaOH solution. [M(NaOH) = 40 g·mol⁻¹]
4.4 (3) Calculate [H₃O⁺] in the NaOH solution at 25°C.
4.5 (3) Explain the difference between a dilute acid and a strong acid.
4.6 (1) From the table (Methyl orange 3.1–4.4; Bromothymol blue 6.0–7.6; Phenolphthalein 8.3–10.0), which indicator is most suitable for Reaction B?
4.7 (2) Define neutralisation/equivalence point.
4.8 (4) Calculate the moles of EXCESS HCl neutralised by the NaOH.
4.9 (6) Calculate the initial concentration of the dilute HCl. [M(CaCO₃) = 100 g·mol⁻¹]
4.10.1 (2) Define hydrolysis of a salt.
4.10.2 (5) Explain whether CaCl₂ solution is acidic, basic, or neutral. Include hydrolysis equations.
✓ Full Memo — DSG 2021
4.1 Standard solution. [1]
4.2 NaOH(s) → Na⁺(aq) + OH⁻(aq) — single arrow, correct phase symbols. [3]
4.3 M(NaOH) = 40. n = m/M = 2/40 = 0,05 mol. V = 250/1000 = 0,25 dm³. c = n/V = 0,05/0,25 = 0,2 mol·dm⁻³ [4]
4.4 [OH⁻] = 0,2 mol·dm⁻³. Kw = [H₃O⁺][OH⁻]. [H₃O⁺] = 1×10⁻¹⁴ ÷ 0,2 = 5×10⁻¹⁴ mol·dm⁻³ [3]
4.5 Dilute = little solute in a lot of solvent (refers to concentration). Strong = ionises almost completely (refers to degree of ionisation). These are independent properties. [3]
4.6 Bromothymol blue. (Reaction B: strong acid + strong base → neutral salt, pH ≈ 7.) [1]
4.7 The point where acid and base have reacted so neither is in excess. [2]
4.8 n(NaOH) = cV = 0,2 × 25/1000 = 0,005 mol. From reaction B ratio 1:1, n(excess HCl) = 0,005 mol. [4]
4.9 n(CaCO₃) = 1,5/100 = 0,015 mol. From reaction A ratio 1:2, n(HCl reacted with CaCO₃) = 0,03 mol. n(HCl total) = 0,03 + 0,005 = 0,035 mol. c(HCl) = 0,035/0,05 = 0,7 mol·dm⁻³ [6]
4.10.1 The reaction of an ion (from a salt) with water. [2]
4.10.2 Ca²⁺ is the conjugate acid of the weak base Ca(OH)₂ and will accept OH⁻ from water: Ca²⁺ + 2H₂O → Ca(OH)₂ + 2H⁺. This increases [H⁺] → solution is slightly acidic (pH < 7). Cl⁻ from strong acid HCl does NOT hydrolyse. [5]
Waverley 2021
Q5 — Ka of Ethanoic Acid (ICE Table) · NaOH Titration · Hydrolysis · Indicators
0,0415 mol of ethanoic acid dissolved in 1 dm³. Equilibrium [CH₃COOH] = 0,0410 mol·dm⁻³. NaOH (from burette) added to 25 cm³ of 0,41 mol·dm⁻³ CH₃COOH. End point at 22,6 cm³ NaOH.
5.1 (2) Define ionisation.
5.2 (6) Use an ICE/RICE table to calculate Ka of ethanoic acid.
5.3.1 (2) Write an equation for the reaction between NaOH and CH₃COOH.
5.3.2 (4) Calculate the concentration of the NaOH solution. [n(CH₃COOH) = 0,0103 mol]
5.3.3 (2) Calculate [H₃O⁺] in the NaOH solution at 25°C.
5.3.4 (3) Would CH₃COOH or NaOH conduct electricity better? Explain.
5.4 (4) The salt CH₃COONa is formed. Use hydrolysis to explain why this salt solution is basic.
5.5.1–5.5.3 (3) For each titration, decide which indicator gives a precise endpoint: Acid Q (weak) + Base X (strong); Acid P (strong) + Base X (strong); Acid P (strong) + Base Y (weak).
✓ Full Memo — Waverley 2021
5.1 Ionisation is the reaction of a molecular substance with water to produce ions. [2]
5.2 — RICE Table:
| Row | CH₃COOH | CH₃COO⁻ | H₃O⁺ |
|---|---|---|---|
| R (ratio) | 1 | 1 | 1 |
| I (initial) | 0,0415 | 0 | 0 |
| C (change) | −0,0005 | +0,0005 | +0,0005 |
| E (equilibrium) | 0,0410 | 0,0005 | 0,0005 |
Ka = [CH₃COO⁻][H₃O⁺] / [CH₃COOH] = (0,0005)² / 0,0410 = 6,1 × 10⁻⁶ [6]
5.3.1 NaOH + CH₃COOH → CH₃COONa + H₂O [2: reactants + products]
5.3.2 Ratio 1:1. n(NaOH) = 0,0103 mol. c = 0,0103/0,0226 = 0,456 mol·dm⁻³ [4]
5.3.3 [H₃O⁺] = 1×10⁻¹⁴ / 0,456 = 2,2×10⁻¹⁴ mol·dm⁻³ [2]
5.3.4 NaOH conducts better — it is a strong base, fully dissociates → many ions. CH₃COOH is weak, partially ionises → few ions. More ions = better conductor. [3]
5.4 CH₃COO⁻ + H₂O ⇌ CH₃COOH + OH⁻. Weak acid forms and does not fully ionise → H₃O⁺ removed from solution → equilibrium in water shifts right → [OH⁻] increases → basic (pH > 7). [4]
5.5.1 Q (weak) + X (strong) → endpoint pH > 7 → Phenolphthalein. [1]
5.5.2 P (strong) + X (strong) → endpoint pH = 7 → Neither (neither indicator covers pH 7 exactly). [1]
5.5.3 P (strong) + Y (weak) → endpoint pH < 7 → Methyl orange. [1]
IEB 2021
Q4+Q5 — Propanoic Acid Ka · Ba(OH)₂ Titration · Precision vs Accuracy
Propanoic acid CH₃CH₂COOH: Ka = 1,34×10⁻⁵. 0,32 mol·dm⁻³ standard solution prepared in 500 cm³ flask. Titration: 20 cm³ of propanoic acid vs Ba(OH)₂. Average V(Ba(OH)₂) = 0,01894 dm³. Calculated c(Ba(OH)₂) = 0,17 mol·dm⁻³. Actual concentration = 0,18 mol·dm⁻³.
4.1 (2) Define concentration.
4.2 (4) What mass of propanoic acid was needed? [M = 74 g·mol⁻¹]
4.3 (1) Why is propanoic acid a WEAK acid?
4.4 (2) Write the Ka expression for propanoic acid.
4.5 (4) Show using an ICE table that [H₃O⁺] = 2,06×10⁻³ mol·dm⁻³.
4.6 (3) Determine [OH⁻] in the propanoic acid solution.
5.1.1 (1) Identify the mistake in EITHER Step 2 (pipette rinsed with water) OR Step 3 (meniscus above the mark).
5.1.2 (3) Explain how the mistake affects the calculated concentration of Ba(OH)₂.
5.2.1 (2) Define neutralisation.
5.2.2 (3) Write a balanced equation for 2CH₃CH₂COOH + Ba(OH)₂ →
5.2.3 (2) Estimate the pH at the end point of this titration.
5.2.4 (5) Show how the concentration of Ba(OH)₂ = 0,17 mol·dm⁻³.
5.2.5(a) (2) Is Sonali's data PRECISE or IMPRECISE? Give a reason.
5.2.5(b) (2) Is Sonali's data ACCURATE or INACCURATE? Give a reason.
✓ Full Memo — IEB 2021
4.1 The amount of moles of solute per unit volume of solution. [2]
4.2 n = cV = 0,32 × 0,5 = 0,16 mol. m = nM = 0,16 × 74 = 11,84 g [4]
4.3 It ionises only partially in solution (low Ka / low concentration of products compared to reactants). [1]
4.4 Ka = [CH₃CH₂COO⁻][H₃O⁺] / [CH₃CH₂COOH] [2]
4.5 — ICE Table (concentrations):
| CH₃CH₂COOH | CH₃CH₂COO⁻ | H₃O⁺ | |
|---|---|---|---|
| Initial | 0,32 | 0 | 0 |
| Change | −x | +x | +x |
| Equil. | 0,32−x | x | x |
1,34×10⁻⁵ = x²/(0,32−x) → x = [H₃O⁺] = 2,06×10⁻³ mol·dm⁻³ ✓ [4]
4.6 [OH⁻] = 1×10⁻¹⁴ / 2,06×10⁻³ = 4,85×10⁻¹² mol·dm⁻³ [3]
5.1.1 Step 2: pipette rinsed with water instead of propanoic acid. OR Step 3: bottom of meniscus not on the mark (above the mark = less acid measured). [1]
5.1.2 Either error decreases the volume/concentration of propanoic acid → smaller volume of Ba(OH)₂ needed to neutralise → calculated concentration of Ba(OH)₂ is higher than actual. [3]
5.2.1 The point where acid and base have reacted so neither is in excess. [2]
5.2.2 2CH₃CH₂COOH + Ba(OH)₂ → (CH₃CH₂COO)₂Ba + 2H₂O [3]
5.2.3 pH ≈ 9 (any basic pH 8–10 gets both marks; weak acid + strong base → basic salt → pH > 7). [2]
5.2.4 nₐ = cₐVₐ = 0,32 × 0,020 = 0,0064 mol. From equation ratio 2:1, n(Ba(OH)₂) = 0,0064/2 = 0,0032 mol. c(Ba(OH)₂) = 0,0032/0,01894 = 0,17 mol·dm⁻³ ✓ [5]
5.2.5(a) IMPRECISE — the readings (0,01691; 0,01896; 0,02095) are NOT close to each other. [2]
5.2.5(b) ACCURATE — 0,17 mol·dm⁻³ is close to the actual value of 0,18 mol·dm⁻³. [2]
⚖️ Chemical Equilibrium — 2021
Heron Bridge 2021
Q4 — SO₂/SO₃ Contact Process · Kc Calculation · Le Chatelier · Temperature & Kc
The Contact Process reaction: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH < 0. Temperature kept between 370°C and 550°C. At equilibrium (400°C): Volume = 200 dm³, initial SO₂ = 50 mol, equilibrium SO₃ = 22 mol, Kc = 7,328.
📊 Rate vs Time Graph (SO₂ reacting with excess O₂; temperature increased to 500°C at t₂)
t₀→t₁: rates converge to equilibrium (dynamic equilibrium reached). At t₂, temperature jumps to 500°C — both rates increase sharply (higher KE), but since the forward reaction is exothermic, the reverse (endothermic) rate increases MORE, shifting equilibrium left. By t₃, a new (higher) equilibrium rate is reached with less SO₃.
4.1 (1) What is represented by the double arrow (⇌) in the equation?
4.2 (2) The reaction is in a closed system. Define a closed system.
4.3 (2) In terms of reaction rate, explain why the temperature must not decrease below 370°C.
4.4 (3) Explain the effect on the yield of SO₃ if the temperature increases beyond 550°C.
4.5.1 (2) Calculate the moles of SO₂ present at equilibrium.
4.5.2 (3) Write the Kc expression AND calculate the concentration of oxygen at equilibrium. (3 decimal places)
4.5.3 (3) Determine the mass of oxygen that was used for this reaction.
4.6.1 (1) The graph shows rate vs time. Name the phenomenon observed between t₁ and t₂ (when the system was at equilibrium).
4.6.2 (4) At t₂ the temperature is increased to 500°C. Explain the changes in the graph using Le Chatelier's Principle.
4.6.3 (3) Explain the effect, if any, on the Kc value due to the temperature change at t₂.
✓ Full Memo — Heron Bridge 2021
4.1 Reversible reaction. [1]
4.2 A closed system is one in which mass is conserved inside the system, but energy can enter or leave the system. [2]
4.3 The rate of both the forward and reverse reactions will decrease if temperature drops below 370°C. The reaction would become too slow to be economically viable. [2]
4.4 The forward reaction is exothermic. If temperature increases beyond 550°C, the reverse (endothermic) reaction is favoured. The rate of the reverse reaction increases more than the forward. The yield of SO₃ will DECREASE. [3]
4.5.1 From equation: n(SO₂) : n(SO₃) = 2:2 = 1:1. So 22 mol SO₂ reacted. n(SO₂) at equilibrium = 50 − 22 = 28 mol [2]
4.5.2 Kc expression: Kc = [SO₃]² / ([SO₂]²[O₂])
Substituting: [SO₃] = 22/200 = 0,11; [SO₂] = 28/200 = 0,14
7,328 = (0,11)² / ((0,14)² × [O₂]) → [O₂] = 0,084 mol·dm⁻³ [3]
4.5.3 n(O₂) at equilibrium = cV = 0,084 × 200 = 16,8 mol. Ratio SO₂:O₂:SO₃ = 2:1:2, so n(O₂) reacted = 22/2 = 11 mol. Total initial O₂ = 16,8 + 11 = 27,8 mol. Mass = nM = 27,8 × 32 = 889,6 g [3]
4.6.1 Dynamic chemical equilibrium is reached in the system. [1]
4.6.2 The increase in temperature acts as a stress on the equilibrium. Both forward and reverse reaction rates increase. Le Chatelier's Principle states the reverse (endothermic) reaction is favoured to absorb the excess heat. Heat is removed from the system to re-establish equilibrium. [4]
4.6.3 Temperature change DOES affect Kc. Since increasing temperature favoured the reverse reaction, concentration of products decreases and reactants increases. Therefore Kc DECREASES after the new equilibrium is established. [3]
Hilton College 2021
Q4 — COCl₂ ⇌ CO + Cl₂ · Graph Reading · Identifying Stress · Kc Calculation · Deducing Stress from Kc
COCl₂(g) ⇌ CO(g) + Cl₂(g) ΔH = +107,6 kJ·mol⁻¹. Kc at 395°C = 1,2×10³. Variable volume container. Already at equilibrium at t=0. At t=5 min: Cl₂ concentration increases sharply. At t=15 min: all concentrations drop simultaneously. At t=25 min: new equilibrium [COCl₂]=0,009; [CO]=0,40; [Cl₂]=0,52 mol·dm⁻³.
📊 Concentration vs Time Graph (0–15 min)
At t=5 min, [Cl₂] jumps up suddenly (Cl₂ added) — the reverse reaction is favoured, so [CO] and [Cl₂] decrease back toward equilibrium while [COCl₂] increases. At t=15 min, ALL concentrations drop together (pressure decrease / volume increase).
📊 Q4.7 — Rate vs Time Graphs (choose A, B, C or D for 10–25 min)
Decreasing pressure (volume increase) at t=15 causes BOTH rates to drop suddenly (less concentrated = fewer collisions), then the forward rate recovers further because the equilibrium shifts toward more gas moles (right side has 2 mol vs 1 mol on left) — this matches graph B.
4.1 (2) Explain what dynamic chemical equilibrium refers to.
4.2 (1) Identify the stress that occurred at 5 minutes.
4.3 (2) State Le Châtelier's Principle.
4.4 (3) Use Le Châtelier's Principle to fully explain, in bullet points, the changes between 5 and 10 minutes on the graph.
4.5 (3) Identify the disturbance at 15 minutes and fully explain your answer.
4.7 (2) Which rate vs time graph (A, B, C or D) is most correct for the period between 10 and 25 minutes? (Forward reaction = solid line; reverse = dashed.)
4.8.1 (4) Calculate Kc at the new equilibrium (after t=25 min).
4.8.2 (5) Use your Kc answer to identify the EXACT nature of the stress at 25 minutes. Explain fully.
✓ Full Memo — Hilton College 2021
4.1 Dynamic chemical equilibrium refers to a reversible reaction in which the forward and reverse reactions are taking place at the same rate, and the concentrations of reactants and products are constant. [2]
4.2 Increase in concentration of Cl₂ (Cl₂ was added). [1]
4.3 When an external stress (change in pressure, temperature or concentration) is applied to a system in dynamic chemical equilibrium, the equilibrium point will change in such a way as to counteract the stress. [2]
4.4 Stress = ↑[Cl₂]. System counteracts by consuming Cl₂ → reverse reaction is favoured (equilibrium shifts LEFT). [Cl₂] and [CO] decrease. [COCl₂] increases. Until new equilibrium. [3 — no mark for restating stress from 4.2]
4.5 At t=15 min, concentrations of ALL gaseous substances dropped simultaneously. This indicates the VOLUME of the container increased. Therefore the disturbance is a DECREASE IN PRESSURE. [3]
4.7 Answer: B. After the pressure drop (t=15), both rates drop suddenly then the forward rate climbs faster (more gas moles on right side → equilibrium shifts right) until rates equalise at new equilibrium. [2]
4.8.1 Kc expression must be stated: Kc = [CO][Cl₂] / [COCl₂]
Kc = (0,40)(0,52) / (0,009) = 23,1 [4 — expression must be written to get marks]
4.8.2 New Kc (23,1) < original Kc (1200). Kc has DECREASED. Since only temperature changes Kc, the stress was a change in temperature. Lower Kc means fewer products / more reactants at equilibrium. The reverse reaction was favoured. The reverse reaction is EXOTHERMIC (ΔH forward = +107,6, so reverse is −107,6). A decrease in temperature favours the exothermic reaction. Therefore the stress was a DECREASE IN TEMPERATURE. [5]
St John's College 2021
Q5 — Ostwald Process (4NH₃ + 5O₂) · ICE Table Kc · Closed System · Exo/Endothermic from Graph
4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g). 2 dm³ container. Kc between t=0 and t=5 min is 1,00×10⁻⁵. At t=5 min a stress is applied. Equilibrium [NO] between t=8–10 min = 1,7 mol·dm⁻³. From graph at t=0: [NH₃]=3,0; [O₂]=2,5; [NO]=1,5; [H₂O]=0,5 mol·dm⁻³.
📊 Concentration vs Time (Ostwald Process — first equilibrium 0–5 min, stress at t=5, new equilibrium by t=8)
0–5 min: system at first equilibrium (all lines flat). At t=5, a temperature increase is applied — NH₃ and O₂ decrease while NO and H₂O increase, meaning the FORWARD reaction was favoured by the temperature increase. A new equilibrium is reached by t=8 with more products than before → Kc increased → forward reaction is ENDOTHERMIC.
5.1 (2) What is meant by closed system?
5.2 (7) If the equilibrium [NO] between t=8–10 min is 1,7 mol·dm⁻³, calculate Kc for this period using an ICE table. Container volume = 2 dm³.
5.3 (2) State Le Châtelier's Principle.
5.4 (3) If the stress at t=5 min was a temperature increase, determine whether the forward reaction is exothermic or endothermic. Use your Kc answer from 5.2 to support.
5.5 (1) If a catalyst was also added at t=5 min, how would the graph change between t=5 and t=8 min?
✓ Full Memo — St John's College 2021
5.1 A system where energy can enter or exit but matter cannot. [2]
5.2 — ICE Table (moles in 2 dm³ container):
| Row | NH₃ | O₂ | NO | H₂O |
|---|---|---|---|---|
| Initial (mol) | 6,00 | 5,00 | 3,00 | 1,00 |
| Change (mol) | −0,40 | −0,50 | +0,40 | +0,60 |
| Equilib (mol) | 5,60 | 4,50 | 3,40 | 1,60 |
| Conc (mol·dm⁻³) | 2,80 | 2,25 | 1,70 | 0,80 |
Kc = [NO]⁴[H₂O]⁶ / ([NH₃]⁴[O₂]⁵)
Kc = (1,70)⁴(0,80)⁶ / ((2,80)⁴(2,25)⁵) = 6,18 × 10⁻⁴ [7]
5.3 When a stress is applied to a system at equilibrium, it responds in a direction that counteracts that stress. [2]
5.4 ENDOTHERMIC. The temperature increase resulted in MORE products being produced — Kc increased from 1,00×10⁻⁵ to 6,18×10⁻⁴. A higher Kc means the forward reaction was favoured by the temperature increase. According to Le Chatelier's, increasing temperature favours the endothermic reaction. Therefore the forward reaction is ENDOTHERMIC. [3]
5.5 It will reach equilibrium faster (in a shorter period of time). The shape of the curves between t=5 and t=8 would be steeper. [1]
IEB 2021
Q3 — Haber Process · NH₃ Yield Graph · Industrial Conditions · Catalyst · Removing Product
N₂(g) + 3H₂(g) ⇌ 2NH₃(g). Graph shows % NH₃ yield vs temperature at pressures 50, 100, 200, 400 atm. Industry operates at 450°C and 200 atm. At 0°C yield is >90%. At 450°C, 200 atm: yield ≈ 16%.
📊 % NH₃ Yield vs Temperature at Different Pressures
Every curve slopes DOWN as temperature increases → higher temperature decreases NH₃ yield → reverse reaction favoured by heat → forward reaction is EXOTHERMIC. Every curve is HIGHER at greater pressure → higher pressure increases yield (fewer moles of gas on the product side). The marked point shows industry's compromise: only ~16% yield at 450°C/200 atm, far below the >90% possible at 0°C — but at a usable reaction rate.
3.1 (1) Is the forward Haber reaction EXOTHERMIC or ENDOTHERMIC?
3.2 (4) Explain your answer to 3.1 using Le Châtelier's Principle with reference to the graph.
3.3 (1) What is the % yield of NH₃ at 450°C and 200 atm?
3.4 (1) Will high pressures favour production of NH₃? YES or NO.
3.5 (3) The graph shows yield at 0°C is >90%, yet industry uses 450°C. Fully explain why.
3.6 (2) An iron catalyst is used. How does this affect the percentage yield of NH₃? State INCREASE, DECREASE, or NO EFFECT.
3.7 (3) Bongani finds a Kc measured at 472°C and 300 atm. Can he use this for conditions at 450°C and 200 atm? Explain.
3.8 (2) Define open system in chemistry.
3.9 (3) NH₃ is continuously removed by dissolving. With reference to REACTION RATE, explain how this affects the yield of NH₃.
✓ Full Memo — IEB 2021
3.1 EXOTHERMIC. [1]
3.2 From the graph, increasing temperature DECREASES the yield of NH₃. An increase in temperature favours the reverse reaction. According to Le Châtelier's Principle, an increase in temperature favours the reaction that decreases temperature — the endothermic reaction. The endothermic reaction is the reverse, meaning the forward reaction is exothermic. [4]
3.3 ≈ 16% (accept 15,5%–16%). [1]
3.4 YES. More pressure → equilibrium shifts toward fewer moles of gas (right side = 2 mol; left = 4 mol) → more NH₃. [1]
3.5 At 0°C the yield is high BUT the reaction rate is extremely slow — not economically viable. At 450°C the yield is lower BUT the rate is fast enough to produce more total NH₃ per unit time. Industry sacrifices some yield to gain an acceptable rate. [3]
3.6 NO EFFECT on percentage yield. The catalyst speeds up both forward and reverse reactions equally, reaching equilibrium faster but at the same equilibrium position. Kc is unchanged. [2]
3.7 NO — Bongani cannot use this Kc value. Kc is temperature-dependent. The temperature at which that Kc was measured (472°C) is different from the industrial temperature (450°C). A different temperature means a different Kc value. [3]
3.8 An open system is one in which both energy AND matter can be exchanged between the system and its surroundings. [2]
3.9 Continuously removing NH₃ (product) decreases the reverse reaction rate. The forward reaction rate is now greater than the reverse. The forward reaction is favoured. More NH₃ is produced — the yield increases. [3]
Every definition tested across the 2019–2023 papers. The highlighted keywords are what examiners specifically mark. Learn the exact wording — paraphrasing costs marks.
⚛️ Bonding & Structure
Intramolecular bond
A force of attraction between atoms within a molecule.
2 marks — keywords: 'within' + 'atoms' + 'molecule'
Intermolecular force
A force of attraction between molecules (or between separate particles).
2 marks — 'between' + 'molecules'
Ionic bond
The electrostatic force of attraction between oppositely charged ions.
2 marks
Covalent bond
The sharing of at least one pair of electrons by two non-metal atoms.
2 marks
Electronegativity
A measure of the tendency of an atom to attract a bonding pair of electrons towards itself.
2 marks
⚡ Reaction Rates
Reaction rate
The change in concentration (or amount) of a reactant or product per unit time.
2 marks — 'change in concentration' + 'per unit time'
Activation energy (Ea)
The minimum energy required for a reaction to occur / for reactant particles to react.
2 marks
Catalyst
A substance that increases the reaction rate without being permanently consumed in the reaction. It lowers the activation energy.
2 marks
⚖️ Equilibrium
Dynamic equilibrium
A state where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant (not necessarily equal).
2 marks — DO NOT say 'equal concentrations'
Le Chatelier's Principle
When a stress is applied to a system at equilibrium, the system will shift in a direction that opposes/counteracts that stress, until a new equilibrium is established.
2 marks
Reversible reaction
A reaction in which the products can react to reform the original reactants under the same conditions.
2 marks
🧪 Acids & Bases
Brønsted-Lowry acid
A substance that donates a proton (H⁺) to another substance.
2 marks
Brønsted-Lowry base
A substance that accepts a proton (H⁺) from another substance.
2 marks — "all or nothing"
Strong acid
An acid that ionises completely in solution. All molecules dissociate into ions.
2 marks — 'ionises' + 'completely'
Weak acid
An acid that ionises partially/incompletely in solution. Only a fraction of molecules dissociate.
2 marks
Ionisation
The reaction of a molecular substance with water to produce ions.
2 marks — tested in Waverley 2021
Diprotic / Polyprotic acid
An acid that can donate two (or more) protons (H⁺) per molecule. e.g. H₂SO₄, (COOH)₂. Polyprotic = more than one.
1 mark
Standard solution
A solution of known/accurately known concentration.
2 marks
Equivalence point / Neutralisation point
The point in a titration where the acid and base have reacted in stoichiometric amounts — neither is in excess.
2 marks — 'stoichiometric' OR 'neither in excess'
Hydrolysis of a salt
The reaction of an ion (from a salt) with water.
2 marks — tested in every 2021 paper
Concentration
The amount of moles of solute per unit volume of solution.
2 marks — IEB 2021
Precise data
Results that are close to each other (close to the average). Precision is about reproducibility, not correctness.
2 marks — IEB 2021
Accurate data
Results that are close to the true/actual value. Accuracy is about correctness, not reproducibility.
2 marks — IEB 2021
⚖️ Chemical Equilibrium — Extra Definitions (2021)
Closed system
A system in which mass is conserved (matter cannot enter or leave) but energy can enter or leave.
2 marks — tested Heron Bridge & St John's 2021
Open system
A system in which both energy AND matter can be exchanged between the system and its surroundings.
2 marks — IEB 2021
Dynamic equilibrium (full version)
A reversible reaction in which the forward and reverse reactions are taking place at the same rate, and concentrations of reactants and products remain constant.
2 marks — Hilton 2021 exact wording
⚖️ What Changes Kc? — Quick Reference Card
🔋 Electrochemistry
Oxidation
The loss of electrons by an atom, ion, or molecule (or increase in oxidation number).
1 mark
Reduction
The gain of electrons by an atom, ion, or molecule (or decrease in oxidation number).
1 mark
Electrolyte
A substance that conducts electricity in the molten or aqueous state due to the presence of mobile ions.
2 marks
🎨 Indicator Quick Reference
⚡ Rule for choosing an indicator
Strong acid + Strong base → pH 7 at endpoint → Bromothymol blue (6.0–7.6). Strong acid + Weak base → pH < 7 → Methyl orange (3.1–4.4). Weak acid + Strong base → pH > 7 → Phenolphthalein (8.3–10.0). Weak acid + Weak base → No single indicator works precisely.
| Indicator | pH Range | Colour Change (acid→base) | Use for |
|---|---|---|---|
| Methyl orange | 3,1 – 4,4 | Red → Yellow | Strong acid + Weak base |
| Methyl red | 3,1 – 4,4 | Red → Yellow | Strong acid + Weak base |
| Bromothymol blue | 6,0 – 7,6 | Yellow → Blue | Strong acid + Strong base |
| Phenolphthalein | 8,3 – 10,0 | Colourless → Pink | Weak acid + Strong base |
These strategies apply across every question in P2. Master these habits and you stop losing marks for content you actually know.
Read the question verb twice
Note whether it says define, explain, calculate, state or describe. "State" = one line. "Explain" = reason + consequence. "Calculate" = every step shown.
Marks = minimum answer points
If a question is 4 marks, you need 4 marking points. Count them before moving on. One long sentence does not earn 4 marks — examiners mark individual points.
Calculations: never skip a step
Formula first → substitute → calculate. Even with a wrong final answer, you earn method marks. One skipped step can cost 2 marks.
Titration calculations: always state the molar ratio
Write the ratio (e.g. "NaOH : HCl = 1:1 ∴ n(NaOH) = n(HCl)") before using it. The ratio mark is often worth 1 mark on its own in the memo.
Balance atoms AND charge in equations
For ionic equations check total charge equals on both sides. For half-reactions, electrons must cancel when combined. One unbalanced atom = 0 marks for the whole equation.
Use the data sheet every time
Look up E° values every single time — one wrong value from memory loses the entire calculation mark. The table is there for a reason.
Specify 25°C when using Kw = 1×10⁻¹⁴
Kw only equals 1×10⁻¹⁴ at 25°C. Questions that ask for [H₃O⁺] or [OH⁻] always specify the temperature — make sure you reference it in your answer.
Hydrolysis questions: always write the equation
When explaining why a salt solution is acidic or basic, you must write the hydrolysis equation AND explain what it does to [H₃O⁺] or [OH⁻]. The equation itself is usually worth 1–2 marks.
What the question verb means
State / Give / Name
One-line answer. No explanation needed.
Define
Use the textbook definition. Key words are marked individually.
Explain
Give the reason AND the consequence. "Because… therefore…"
Describe
What do you observe? Be specific. Include what changes and in what direction.
Calculate
Formula → substitution → answer with units. All three steps visible.
Write an equation
Balance atoms AND charge. State symbols only if the question asks.
Predict and explain
State your prediction first, then justify with chemistry reasoning.
Justify / Substantiate
Prove your answer is correct using calculations or chemical reasoning.
Titration calculation — checklist every time
1
Write the balanced equation for the reaction
2
State the molar ratio from the equation
3
Calculate moles of the known substance (n = cV, remember to convert cm³ to dm³)
4
Use the molar ratio to find moles of the unknown
5
Calculate concentration c = n/V (or mass = nM)
6
Write the final answer to 2 decimal places with correct units