P2 Chemistry — Topic Overview
Six topics examined in Paper 2. Tap any card to see key subtopics and examiner focus areas.
Organic Chemistry
~35 marks
Nomenclature, functional groups, homologous series, physical properties (IMFs and boiling points), reactions (addition, elimination, substitution, esterification, combustion).
Acids & Bases
~25 marks
Arrhenius and Brønsted-Lowry models, conjugate pairs, pH calculations, buffer solutions, indicators, neutralisation and titration calculations.
Chemical Equilibrium
~25 marks
Dynamic equilibrium, Le Chatelier's principle, Kc expression and calculations, factors affecting equilibrium (concentration, temperature, pressure).
Electrochemistry
~25 marks
Galvanic and electrolytic cells, standard reduction potentials, cell notation, Ecell calculations, electroplating, electrolysis of water and brine.
Reaction Rates
~15 marks
Factors affecting rate, collision theory, activation energy, potential energy diagrams (catalysed vs uncatalysed), rate expressions and calculations.
Quantitative Chemistry
~25 marks
Molar calculations, stoichiometry, limiting reagents, percentage yield, percentage purity, concentration calculations (c = n/V), gas volumes at STP, titrations.
Past Paper Practice
IEB school papers (2018–2021) with full memos and examiner tips. Filter by topic.
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More questions being added. Organic Chemistry has full coverage. Additional questions for Rates, Bonding and Quantitative Chemistry are being added progressively. In the meantime, use the NSC Question Bank → for full topic coverage across all 8 topics.
Organic Chemistry
[19]
Organic Reaction Sequence — Homologous Series & Reaction Types
Consider the following sequence of organic reactions (reactions labelled A–E, molecules labelled II–V):
Molecule II (alkene) ←B→ Molecule III (ethanol: H-C-C-OH) ←A/D→ Molecule II
Molecule II →C→ Molecule V (chloroethane: H-C-C-Cl)
Molecule IV (ethyl propanoate ester) ←B→ Molecule III
Molecule II →E→ CO₂ + H₂O
What is meant by the term homologous series? (2)
Identify the homologous series to which II (alkene), III (alcohol), and IV (ester) belong. (3)
Name the types of reactions labelled A, C, D and E. Provide the specific reaction type where applicable. (4)
Write a balanced chemical equation using molecular formulae for reaction E (combustion of ethene). (2)
Molecule III (ethanol) is completely soluble in water. Explain in terms of intermolecular forces why. (3)
For reaction B to form molecule IV, molecule III reacts with a molecule from a different homologous series. Name that homologous series, give the IUPAC name of that molecule, and give the IUPAC name of molecule IV. (5)
Model Answers
a
A series of similar compounds with the same functional group and same general formula, in which each member differs from the previous by a single CH₂ unit.
b
II: Alkenes | III: Alcohols | IV: Esters
c
A: Hydration (addition) | C: Substitution (halogenation) | D: Dehydration (elimination) | E: Combustion
d
C₂H₄ + 3O₂ → 2CO₂ + 2H₂Oe
Ethanol has an –OH functional group. The O–H bond is polar, so ethanol can form hydrogen bonds with water molecules (which also have a polar O–H site). Like dissolves like — both molecules form hydrogen bonds.
f
Homologous series: Carboxylic acids | Other molecule: Propanoic acid | Molecule IV: Ethyl propanoate
⚠ Examiner Tip For reaction type questions: "substitution" alone is not enough — you must say "halogenation" or "nucleophilic substitution" to get the specific-type mark. Likewise for A: "addition" alone gets you the mark here, but "hydration" is the specific term and gains full credit.
Organic Chemistry
[18]
Boiling Points, IMFs & Structural Isomers
The boiling points of five organic compounds are given:
Boiling Points of Organic Compounds
| Letter | Compound | Boiling Point (°C) |
|---|---|---|
| A | Propane | −42,0 |
| B | Pentane | 36,0 |
| C | Methylbutane | 27,8 |
| D | Hexan-1-ol | 157,0 |
| E | Pentanoic acid | 186,0 |
Give a balanced chemical equation for the complete combustion of compound A (propane). (5)
An unknown straight-chain alkane has a boiling point of −0,5°C. Name this alkane using the table. (1)
Compounds B and C are structural isomers. Define structural isomer, state what type of structural isomers they are, and explain why B has a higher boiling point than C. (4+1+4 = 7)
Compounds D and E have the same molar mass. Are they structural isomers? Explain. Then explain the difference in their boiling points. (2+3 = 5)
Model Answers
a
C₃H₈ + 5O₂ → 3CO₂ + 4H₂O (✓✓ reactants, ✓✓ products, ✓ balancing)b
Butane (boiling point of n-butane is −0,5°C — between propane and pentane)
c
Definition: Compounds with the same molecular formula but different structural formulae. Type: Chain isomers. Explanation: Both have London forces. B (pentane, straight chain) has a larger contact surface area than C (methylbutane, branched) → larger temporary dipoles → stronger London forces → more energy required to overcome them → higher boiling point.
d
Not structural isomers — they have different molecular formulae (D: C₆H₁₄O; E: C₅H₁₀O₂). Boiling points: Both have hydrogen bonding. E (carboxylic acid) has two polar sites (C=O and O–H) and can form stronger/more hydrogen bonds than D (alcohol, one O–H site) → more energy to overcome → higher boiling point.
⚠ Examiner Tip When comparing boiling points between London-force molecules: always state (1) same IMF type, (2) which is stronger and why (surface area/molecular mass), (3) energy required. Missing any step = lose marks.
Organic Chemistry
[16]
Ester Synthesis, Functional Isomers & Nomenclature
Compound Z is an ester (propyl ethanoate) produced in a condensation reaction between ethanoic acid and propan-1-ol.
Name the homologous series to which compound Z belongs. (1)
Give the IUPAC name of compound Z. (2)
Define structural isomers. Give the IUPAC name of the functional isomer of compound Z and name its functional group. (2+2+2 = 6)
Describe a safe way to heat the reactants when making an ester in the laboratory and explain why this method is necessary. (2)
What other reaction condition (besides heating) must be met for esterification to occur, and why is it necessary? (3)
Model Answers
a
Esters
b
Propyl ethanoate
c
Structural isomers: Compounds with the same molecular formula but different structural formulae. Functional isomer: Pentanoic acid. Functional group: Carboxyl group (–COOH)
d
Heat indirectly using a water bath. Reason: the reactants (alcohol and carboxylic acid) are highly flammable — a water bath prevents direct contact with a flame and avoids the risk of ignition/evaporation.
e
Concentrated H₂SO₄ (sulphuric acid) must be added as a catalyst. It acts as a dehydrating agent, removing the water produced and driving the equilibrium towards the ester product (Le Chatelier's principle).
⚠ Examiner Tip "Heat using water bath" is a two-mark answer: name the method AND state the reason (flammable reactants). Writing only "water bath" gets you 1/2.
Organic Chemistry
[27]
Cracking, Reaction Types, IMFs & Boiling Points
A long-chain hydrocarbon C₉H₂₀ undergoes Reaction 1 (high pressure and temperature) to produce hexane and Compound L (propene).
Compound L + HCl → Compound M (CH₃CHClCH₃) [Reaction 2]
Compound M + substance (iv) → Alcohol N + salt (v) [Reaction 3]
For Reaction 1: identify the general type and the specific type of reaction. (2)
Write the balanced combustion reaction for hexane using molecular formulae. (3)
Compound K is a structural isomer of hexane. Give the IUPAC name for one possible isomer. What specific type of isomerism is shown? (3+1)
Give the IUPAC name for Compound L (propene). (1)
For Reaction 2: identify the specific reaction type. What condition is required? (1+1)
For Reaction 3: identify the general type. Draw alcohol N's structure. Identify substance (iv) and salt (v). State the conditions. (1+3+2+2)
Identify the IMF in Compounds L, M and N. Rank them in increasing order of boiling points. Explain which has the highest boiling point. (3+2+2)
Model Answers
a
General: Elimination | Specific: (Thermal) cracking
b
2C₆H₁₄ + 19O₂ → 12CO₂ + 14H₂Oc
Any valid: 2-methylpentane / 3-methylpentane / 2,3-dimethylbutane etc. Type: Chain isomerism
d
Propene
e
Specific type: Hydrohalogenation (hydrochlorination). Condition: Absence of water / anhydrous conditions
f
General type: Substitution. Alcohol N: propan-2-ol (–OH on middle carbon). Substance (iv): NaOH or KOH. Salt (v): NaCl or KCl. Conditions: dilute NaOH (alkali solution) and heat (under reflux).
g
L (propene): London forces | M (chloropropane): dipole-dipole forces | N (propan-2-ol): hydrogen bonds. Order: L → M → N. N has the highest: hydrogen bonding is the strongest IMF, requiring the most energy to overcome, hence the highest boiling point.
⚠ Examiner Tip Ranking boiling points = identify IMF type first. The mark scheme awards a mark for correctly identifying the IMF in each compound before comparing. Don't skip this step.
Organic Chemistry
[20]
Linalool — Functional Groups, Unsaturation & IUPAC Naming
Linalool is an active ingredient in the aroma of lilies. It is a 10-carbon compound containing two C=C double bonds and one –OH group.
Define functional group. (2)
Name the functional groups in linalool. (2)
"Linalool is a saturated hydrocarbon." Do you agree with any part of this statement? Explain. (3)
Linalool has a boiling point of 199°C. Explain why this is substantially higher than water's boiling point. (3)
Compound Y is a related molecule (3-bromo-3,7-dimethylocta-1,6-diene). Give the IUPAC name for compound Y. (4)
Define structural isomer. Draw a chain isomer of Y consisting of a 6-carbon main chain (include both double bonds and bromine). (2+4)
Model Answers
a
An atom or group of atoms that form the centre of chemical activity in a molecule.
b
Double bond (C=C) and hydroxyl group (–OH)
c
Partially agree. It contains carbon and hydrogen → it is a hydrocarbon in that sense. BUT it is not saturated (contains C=C double bonds → unsaturated) and it is not a hydrocarbon (contains an –OH group → it is an alcohol).
d
Linalool has a much higher molecular mass than water → more electrons → greater distortion of electron cloud → stronger London forces. More energy is required to overcome these stronger forces → higher boiling point. (Note: linalool cannot hydrogen-bond as effectively as water because water has more H-bonding sites per unit mass.)
e
3-bromo-3,7-dimethylocta-1,6-diene (marking: −1 for each error)
f
Structural isomers: same molecular formula, different structural formulae. Chain isomer: any valid structure with a 6-carbon main chain, same molecular formula as Y, double bonds and Br correctly placed.
⚠ Examiner Tip For "is it a hydrocarbon?" questions: a hydrocarbon contains C and H only. Any –OH, =O, –Cl etc. makes it NOT a hydrocarbon, regardless of other properties. This is a common mark-losing error.
Organic Chemistry
[20]
Methyl Butanoate Synthesis — Halogenation to Esterification
Methyl butanoate (pineapple smell) is synthesised from alkane X via three steps:
Alkane X →[STEP 1: +Cl₂, heat/sunlight]→ Y →[STEP 2: +NaOH(aq), heat]→ Z (methanol) →[STEP 3: +carboxylic acid, H₂SO₄]→ Methyl butanoate
Name alkane X. (1)
Write a balanced equation for Step 1 using condensed structural formulae. (3)
What specific type of reaction occurs in Step 1? (1)
Explain how the structure of alkane X influences the rate of Step 1. (2)
Define the term homologous series. (2)
To which homologous series does compound Z belong? Name compound Z. (1+1)
For Step 3: name the process; name the functional group of the carboxylic acid; name the carboxylic acid; write a structural equation for Step 3. (1+1+1+4)
State the heating method for Step 3 and give a reason. (2)
Model Answers
a
Methane (CH₄)
b
CH₄ + Cl₂ → CH₃Cl + HClc
Halogenation (chlorination) — specific type of substitution
d
Methane is saturated (only single bonds) so the reaction rate is slow — more energy is required to break the strong C–H bonds compared to unsaturated compounds.
e
A series of similar compounds with the same functional group and same general formula, in which each member differs from the previous by a CH₂ unit.
f
Homologous series: Alcohols. Name: Methanol
g
Process: Esterification. Functional group: Carboxyl (–COOH). Carboxylic acid: Butanoic acid. Equation: CH₃OH + C₃H₇COOH ⇌ C₃H₇COOCH₃ + H₂O (with H₂SO₄ catalyst)
h
Water bath. Reason: alcohol (and ester) are flammable — direct flame contact could cause ignition.
⚠ Examiner Tip For condensed structural formulae in Step 1: the question says "condensed" — you must write CH₄, CH₃Cl, HCl, not full structural diagrams. Using full structural diagrams for a "condensed" question risks losing the formula mark.
Organic Chemistry
[15]
Elimination, Addition & Hydrogenation Reactions
A reaction scheme shows: butan-1-ol ←A/B→ but-1-ene →[H₂, Pt]→ Compound Z (butane). Reaction C: butan-1-ol + HBr → Compound X.
For reaction D (X → but-1-ene): identify general type; name specific type; write structural equation with conditions. (1+1+4)
Draw the structure of compound Z (butane). (3)
Alkene Y → compound Z. Identify general type (addition). Name specific type (hydrogenation). Write balanced combustion equation for Z. (1+1+4)
Model Answers
a
General: Elimination. Specific: Dehydrohalogenation (dehydrobromination). Equation: CH₃CH₂CH₂CH₂Br + NaOH(conc) →[ethanol, heat]→ CH₃CH₂CH=CH₂ + H₂O + NaBr
b
H–C–C–C–C–H with all H atoms shown (4 carbons, all single bonds, CH₃CH₂CH₂CH₃)
c
General: Addition. Specific: Hydrogenation. Combustion:
2C₄H₁₀ + 13O₂ → 8CO₂ + 10H₂O⚠ Examiner Tip For elimination reactions: always state the reagent (conc NaOH), solvent (ethanol/hot KOH in ethanol), and conditions (heat). Missing conditions = losing marks. "Dilute NaOH in water" gives substitution — the examiner knows if you don't specify.
Organic Chemistry
[21]
Ester Preparation, Isomers & Boiling Point Comparison
Molecule IV (ethyl propanoate) is prepared by esterification. It has a functional isomer.
Why must the test tube be heated in a hot water bath, not directly over a Bunsen flame? (2)
What is the function of H₂SO₄ in esterification? (2)
Define isomer. (2)
Draw the structural formula of the functional isomer of ethyl propanoate. Give its IUPAC name. (2+2)
Which isomer (ethyl propanoate or the one in 9.2.4) has the higher boiling point? Explain fully. (3)
Give the IUPAC names of the following: 2,3-dimethylbutane; 2-methyl-2-propanol; 3-methyl-2-pentene. (2+2+2)
Write the molecular formula for 2-methyl-2-propanol. (2)
Model Answers
a
Heat is transferred more gently (avoids overheating). The reactants are flammable — reduces risk of ignition.
b
Acts as a catalyst (and dehydrating agent — removes water, driving equilibrium towards products).
c
Isomers are compounds that have the same molecular formula but different structural formulae.
d
Functional isomer: Pentanoic acid (C₄H₉COOH). Full structural formula showing –COOH group on a 5-carbon chain.
e
Pentanoic acid has the higher boiling point. It has two polar sites (O–H and C=O) and can form strong hydrogen bonds. Ethyl propanoate can only accept hydrogen bonds (no O–H). More energy is required to overcome the stronger IMFs of pentanoic acid → higher boiling point.
f
2,3-dimethylbutane ✓✓ | 2-methyl-2-propanol OR 2-methylpropan-2-ol ✓✓ | 3-methyl-2-pentene ✓✓
g
C₄H₁₀O (or C₄H₉OH)
⚠ Examiner Tip "Functional isomer" means same molecular formula but different functional group — not just different arrangement. An ester and a carboxylic acid with the same formula are functional isomers. A common error is drawing a positional isomer (same functional group) instead.
Acids & Bases
[25]
Ba(OH)₂ vs Oxalic Acid Titration — Strong/Weak, Diprotic & Salt Hydrolysis
Ba(OH)₂ (strong alkali) is titrated with (COOH)₂ (oxalic acid — weak diprotic acid). 25 cm³ of 0,2 mol·dm⁻³ Ba(OH)₂ is in the flask with phenolphthalein indicator. 18,6 cm³ of the acid neutralises the alkali.
What is meant by saying Ba(OH)₂ is a strong alkali? (2)
What is a diprotic acid? (1)
Name the dilute alkali and the piece of apparatus labelled X (burette). (1+1)
What is meant by the equivalence point of the titration? (2)
Write a balanced equation for the neutralisation reaction. (2)
Calculate moles of alkali at the start; hence find the concentration of the acid. (3+3)
State whether [Ba²⁺], [OH⁻] and pH INCREASE, DECREASE or REMAIN CONSTANT while acid is added before the end point. (3)
Complete these ionic equations: NH₄⁺(aq) + H₂O(ℓ) → ? and CH₃COO⁻(aq) + H₂O(ℓ) → ? Name the phenomenon. (4+1)
Model Answers
a
A strong alkali completely dissociates (ionises) in aqueous solution.
b
An acid that can donate two protons (H⁺) per molecule.
c
Alkali: Barium hydroxide. Apparatus X: Burette.
d
The point where an acid and base have reacted so that neither is in excess.
e
Ba(OH)₂ + (COOH)₂ → (COO)₂Ba + 2H₂Of
n(Ba(OH)₂) = cV = 0,2 × 0,025 = 5 × 10⁻³ mol. Mole ratio alkali:acid = 1:1. n(acid) = 5 × 10⁻³ mol. C(acid) = n/V = 5×10⁻³ / 0,0186 = 0,27 mol·dm⁻³
g
[Ba²⁺]: Remains constant (spectator ion, not involved in reaction). [OH⁻]: Decreases (being neutralised by the acid). pH: Decreases (as OH⁻ is consumed).
h
NH₄⁺(aq) + H₂O(ℓ) → NH₄OH + H⁺ | CH₃COO⁻(aq) + H₂O(ℓ) → CH₃COOH + OH⁻. Phenomenon: Hydrolysis of a salt.⚠ Examiner Tip Ba²⁺ is a spectator ion — it does NOT react with OH⁻ or the acid. Its concentration remains constant throughout the titration. Writing that [Ba²⁺] decreases is a very common error.
Acids & Bases
[30]
Strong vs Weak Acids, Polyprotic Acids, Ka & Indicator Choice
Two circuits are set up with 1 mol·dm⁻³ HCl and 1 mol·dm⁻³ CH₃CO₂H as electrolytes. Bulb 1 (HCl) is brighter than Bulb 2 (ethanoic acid).
Define a strong acid. (2)
Fully explain why Bulb 1 is brighter than Bulb 2. (3)
Why is H₃PO₄ polyprotic? Write its full ionisation in water (assume complete ionisation). (1+4)
H₃PO₄ concentration in a soft drink = 1,67 × 10⁻³ mol·dm⁻³. Calculate [H₃O⁺] assuming complete ionisation. Hence find [OH⁻] at 25°C. (2+3)
30 cm³ of 2 mol·dm⁻³ NaOH is titrated with 15 cm³ of H₃PO₄ (3NaOH + H₃PO₄ → Na₃PO₄ + 3H₂O). Calculate [H₃PO₄] and name the salt formed. (5+1)
Define hydrolysis of a salt. Using hydrolysis equations, justify the choice of phenolphthalein (pH 8,4–11) as the indicator. (2+4)
Model Answers
a
A strong acid ionises completely (fully) in aqueous solution.
b
HCl is a strong acid → ionises completely → many free ions in solution → conducts a greater current → brighter bulb. CH₃CO₂H is a weak acid → only partially ionises → few ions → low current → dim bulb.
c
H₃PO₄ can donate more than one proton. Ionisation:
H₃PO₄ + 3H₂O ⇌ 3H₃O⁺ + PO₄³⁻d
Ratio H₃PO₄ : H₃O⁺ = 1:3. [H₃O⁺] = 3 × 1,67 × 10⁻³ = 5,01 × 10⁻³ mol·dm⁻³. Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴. [OH⁻] = 10⁻¹⁴ / 5,01×10⁻³ = 2 × 10⁻¹² mol·dm⁻³
e
n(NaOH) = cV = 2 × 0,030 = 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. Salt: Sodium phosphate (Na₃PO₄).
f
Hydrolysis: reaction of an ion (from a salt) with water. PO₄³⁻ is the conjugate base of the weak acid H₃PO₄ → it hydrolyses water:
PO₄³⁻ + H₂O ⇌ OH⁻ + H₂PO₄⁻. This produces excess OH⁻ → solution at end point is basic (pH > 7). Na⁺ does not hydrolyse. Phenolphthalein (range 8,4–11) changes colour in the basic region → correct choice.⚠ Examiner Tip Indicator selection is linked to salt hydrolysis at the endpoint. If the salt is basic (from strong base + weak acid), use phenolphthalein. If neutral (strong acid + strong base), use either. If acidic (weak base + strong acid), use methyl orange. Always justify by stating the expected pH at the equivalence point.
Acids & Bases
[20]
KOH + H₂SO₄ Titration — Brønsted-Lowry, [OH⁻] Calc & Excess Acid Problem
Reaction: 2KOH(aq) + H₂SO₄(aq) → K₂SO₄(aq) + 2H₂O(l)
0,025 mol·dm⁻³ KOH is titrated with 25 cm³ of 0,010 mol·dm⁻³ H₂SO₄.
Provide the Brønsted-Lowry definition for a base. (2)
H₂SO₄ is a strong acid. What does this mean? (2)
Calculate [OH⁻] in the original H₂SO₄ solution before any KOH is added. (4)
Define equivalence point. (2)
Determine the volume of KOH required to fully neutralise the H₂SO₄. (3)
More acid falls in past the endpoint. Explain how this affects the calculated concentration of the base. (2)
2,5 ml of 0,525 mol·dm⁻³ NaOH and 7,5 ml of 0,355 mol·dm⁻³ HCl are accidentally mixed. Show with calculations whether the resulting solution is acidic or basic. (5)
Model Answers
a
A Brønsted-Lowry base is a proton (H⁺) acceptor.
b
The acid will completely ionise in solution.
c
H₂SO₄ is diprotic: [H₃O⁺] = 2 × 0,010 = 0,020 mol·dm⁻³. Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴. [OH⁻] = 10⁻¹⁴/0,020 = 5,00 × 10⁻¹³ mol·dm⁻³
d
The point where an acid and base have reacted so neither is in excess.
e
n(H₂SO₄) = cV = 0,010 × 0,025 = 2,5 × 10⁻⁴ mol. n(KOH) = 2 × n(H₂SO₄) = 5,00 × 10⁻⁴ mol. V = n/C = 5,00×10⁻⁴/0,025 = 0,020 dm³ = 20 cm³
f
More acid than necessary is recorded as having been used. This gives a higher apparent volume of acid → more moles of acid than actually reacted with the base → calculated concentration of base is higher than the actual concentration.
g
NaOH + HCl → NaCl + H₂O (1:1 ratio). n(NaOH) = 0,525 × 0,0025 = 1,31 × 10⁻³ mol. n(HCl) = 0,355 × 0,0075 = 2,66 × 10⁻³ mol. n(HCl) > n(NaOH) → HCl is in excess → [H₃O⁺] > [OH⁻] → solution is acidic.
⚠ Examiner Tip For H₂SO₄: always double the concentration for [H₃O⁺] since it is diprotic (2 H⁺ per molecule). Forgetting this gives [H₃O⁺] half the correct value — a common error that then cascades into a wrong [OH⁻].
Acids & Bases
[34]
NaOH Standard Solution, Back-Titration & Salt Hydrolysis (CaCl₂)
2 g of pure NaOH is dissolved in 250 cm³ volumetric flask. The NaOH solution is used to neutralise excess HCl remaining after 1,5 g CaCO₃ reacted with 50 cm³ dilute HCl.
Reaction A: 2HCl(aq) + CaCO₃(s) → CaCl₂(aq) + CO₂(g) + H₂O(l)
Reaction B: HCl(aq) + NaOH(aq) → NaCl(aq) + H₂O(l) — 25 cm³ NaOH neutralises excess HCl.
What term describes a "solution of known concentration"? Write the dissociation equation for NaOH. Calculate the concentration of the NaOH solution. (1+3+4)
Calculate [H₃O⁺] in the NaOH solution at 25°C. (3)
Explain the difference between a dilute and a strong acid. (3)
Which indicator suits Reaction B? (Methyl orange pH 3,1–4,4; Bromothymol blue 6,0–7,6; Phenolphthalein 8,3–10,0) (1)
Define the neutralisation/equivalence point. Calculate moles of EXCESS HCl; hence calculate the initial concentration of the HCl. (2+4+6)
Define hydrolysis of a salt. Explain whether a CaCl₂ solution is acidic, basic or neutral (include equations). (2+5)
Model Answers — Key Steps
a
Term: Standard solution. Dissociation:
NaOH(s) → Na⁺(aq) + OH⁻(aq). n(NaOH) = 2/40 = 0,05 mol. V = 0,25 dm³. C = 0,05/0,25 = 0,2 mol·dm⁻³b
[OH⁻] = [NaOH] = 0,2 mol·dm⁻³. Kw = [H₃O⁺][OH⁻] = 10⁻¹⁴. [H₃O⁺] = 10⁻¹⁴/0,2 = 5 × 10⁻¹⁴ mol·dm⁻³
c
Dilute = small amount of solute dissolved in a large volume of solvent. Strong = ionises almost completely in solution. These are independent properties — a strong acid can be dilute or concentrated.
d
Bromothymol blue (Reaction B is strong acid + strong base → equivalence point is pH 7 → Bromothymol blue range 6,0–7,6 includes pH 7).
e
Equivalence point: the point where acid and base have reacted so that neither is in excess. n(NaOH) = 0,2 × 0,025 = 0,005 mol. Ratio NaCl:HCl = 1:1 → n(excess HCl) = 0,005 mol. n(HCl) reacting with CaCO₃ = 1,5/100 × 2 = 0,03 mol. Total n(HCl) = 0,03 + 0,005 = 0,035 mol. C(HCl) = 0,035/0,05 = 0,7 mol·dm⁻³
f
Hydrolysis: reaction of an ion (from a salt) with water. Ca²⁺ is a conjugate acid of the weak base Ca(OH)₂ → accepts OH⁻ from water:
Ca²⁺ + 2H₂O → Ca(OH)₂ + 2H⁺. Cl⁻ does NOT hydrolyse (HCl is a strong acid → Cl⁻ is too stable). ∴ [H⁺] increases → solution is slightly acidic (pH < 7).⚠ Examiner Tip "Dilute" and "strong" are two completely different properties. Dilute describes concentration (how much solute). Strong describes degree of ionisation. A strong acid can be dilute; a weak acid can be concentrated. Mixing these up is a classic 3-mark mistake.
Acids & Bases
[41]
Weak Acid Ka Calculation, ICE Table & Propanoic Acid Titration
Propanoic acid (CH₃CH₂COOH, Ka = 1,34 × 10⁻⁵) is prepared as a 0,32 mol·dm⁻³ standard solution in a 500 cm³ flask at 25°C.
Define concentration. Calculate the mass of propanoic acid needed. (2+4)
Why is propanoic acid considered a weak acid? Write the Ka expression. (1+2)
Using the ICE table below, show that [H₃O⁺] = 2,06 × 10⁻³ mol·dm⁻³. Hence find [OH⁻]. (4+3)
ICE Table — CH₃CH₂COOH ionisation (0,32 mol·dm⁻³)
| CH₃CH₂COOH | + | H₂O | ⇌ | CH₃CH₂COO⁻ | + | H₃O⁺ | |
|---|---|---|---|---|---|---|---|
| I | 0,32 | — | 0 | 0 | |||
| C | −x | — | +x | +x | |||
| E | 0,32 − x | — | x | x |
Ka = x²/(0,32−x) ≈ x²/0,32 = 1,34×10⁻⁵ → x = [H₃O⁺] = 2,06×10⁻³ mol·dm⁻³
Sonali titrates 20 cm³ of the propanoic acid with Ba(OH)₂. The pipette was rinsed with water, not acid — identify the error and explain how it affects the calculated [Ba(OH)₂]. (1+3)
The average volume of Ba(OH)₂ used = 0,01894 dm³. Write the balanced equation; estimate the pH at the end point; calculate [Ba(OH)₂]. (3+2+5)
Waverley: 0,0415 mol ethanoic acid dissolved in 1 dm³ water; at equilibrium [CH₃COOH] = 0,0410 mol·dm⁻³. Use the ICE table below to find Ka. (6)
ICE Table — Ethanoic Acid (CH₃COOH) Ka Calculation
| CH₃COOH | ⇌ | CH₃COO⁻ | + | H₃O⁺ | |
|---|---|---|---|---|---|
| I | 0,0415 | 0 | 0 | ||
| C | −0,0005 | +0,0005 | +0,0005 | ||
| E | 0,0410 | 0,0005 | 0,0005 |
Ka = [CH₃COO⁻][H₃O⁺]/[CH₃COOH] = (0,0005)²/0,0410 = 6,1 × 10⁻⁶
Model Answers — Key Steps
a
Concentration: amount of solute per unit volume of solution. n = cV = 0,32 × 0,5 = 0,16 mol. M(propanoic acid) = 74 g·mol⁻¹. m = 0,16 × 74 = 11,84 g
b
It ionises partially in solution (Ka is small / low [H₃O⁺] relative to acid concentration). Ka = [CH₃CH₂COO⁻][H₃O⁺] / [CH₃CH₂COOH]
c
ICE table: Initial [acid]=0,32; [products]=0. Change: −x, +x, +x. Equil: (0,32−x), x, x. 1,34×10⁻⁵ = x²/(0,32−x) ≈ x²/0,32. x = √(1,34×10⁻⁵ × 0,32) = 2,06 × 10⁻³ mol·dm⁻³. [OH⁻] = 10⁻¹⁴/2,06×10⁻³ = 4,85 × 10⁻¹² mol·dm⁻³
d
Error: pipette rinsed with water dilutes the acid → less propanoic acid in flask → less Ba(OH)₂ needed to neutralise → smaller recorded volume → calculated [Ba(OH)₂] is higher than actual.
e
Equation:
2CH₃CH₂COOH + Ba(OH)₂ → (CH₃CH₂COO)₂Ba + 2H₂O. End point pH: ~9 (basic salt from weak acid + strong base). n(acid) = 0,32 × 0,020 = 0,0064 mol. Ratio acid:Ba(OH)₂ = 2:1. n(Ba(OH)₂) = 0,0032 mol. C = 0,0032/0,01894 = 0,17 mol·dm⁻³f
Δ[CH₃COOH] = 0,0415 − 0,0410 = 0,0005 mol (ionised). ICE: E row = 0,0410, 0,0005, 0,0005. Ka = (0,0005)²/0,0410 = 2,5×10⁻⁴/0,0410 = 6,1 × 10⁻⁶
⚠ Examiner Tip ICE table method: the Change row always has −x for reactants and +x for products. The Equilibrium row is Initial + Change. For weak acids, if Ka is very small, approximate: (initial − x) ≈ initial. Always verify this assumption is valid (x must be <5% of initial).
Equilibrium
[24]
Contact Process (SO₂ + O₂ ⇌ SO₃) — Kc Calculation, Le Chatelier & Temperature Effect
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH < 0 (Contact Process for H₂SO₄)
Container volume = 200 dm³ | Initial SO₂ = 50 mol | Equilibrium SO₃ = 22 mol | Kc at 400°C = 7,328
Rate vs Time — SO₂/SO₃ System
Forward reaction (solid green) starts fast as SO₂ and O₂ react; reverse reaction (dashed blue) starts slow as SO₃ builds up. At t₁ both rates are equal — equilibrium reached. At t₂ temperature is raised: both rates instantly increase, but because the reaction is exothermic (ΔH < 0), the reverse (endothermic) reaction is favoured — it increases more. Both rates re-equalise at t₃ at a higher level, with lower [SO₃] yield.
What does the double arrow represent? Define a closed system. (1+2)
Why is the temperature kept above 370°C? (2)
Explain the effect on SO₃ yield if temperature rises above 550°C. (3)
Calculate moles of SO₂ at equilibrium. Write the Kc expression and calculate [O₂] at equilibrium (to 3 decimal places). (2+3)
Determine the mass of O₂ initially present. (3)
At t₂ temperature is raised to 500°C. Explain changes using Le Chatelier's principle. Does Kc change? Explain. (4+3)
Model Answers
a
Double arrow = reversible reaction. Closed system: one in which mass is conserved inside the system but energy can enter or leave.
b
Below 370°C both the forward and reverse reaction rates decrease too much — the rate of reaching equilibrium becomes too slow to be economical.
c
The forward reaction is exothermic. If T rises above 550°C, Le Chatelier's principle → reverse (endothermic) reaction favoured → more SO₃ decomposes → yield of SO₃ decreases.
d
Ratio SO₂:SO₃ = 2:2 = 1:1, so 22 mol SO₃ formed → 22 mol SO₂ consumed. n(SO₂)eq = 50 − 22 = 28 mol. Kc = [SO₃]²/([SO₂]²[O₂]). 7,328 = (22/200)²/((28/200)² × [O₂]). [O₂] = 0,084 mol·dm⁻³
e
n(O₂ unreacted) = 0,084 × 200 = 16,8 mol. n(O₂ reacted) = 22/2 = 11 mol (ratio SO₂:O₂ = 2:1). Total n(O₂) initial = 16,8 + 11 = 27,8 mol. m = 27,8 × 32 = 889,6 g
f
↑T acts as a stress. Le Chatelier: reverse (endothermic) reaction favoured to oppose heat increase → SO₃ concentration decreases, SO₂ and O₂ increase → new equilibrium at higher T. Kc changes: since ↑T favoured reverse reaction, [products] decrease and [reactants] increase → Kc value decreases.
⚠ Examiner Tip Kc only changes when temperature changes — not when concentration or pressure changes. Many learners incorrectly state Kc changes when you add a reactant. It doesn't. Temperature is the only factor that changes Kc.
Equilibrium
[26]
Phosgene Equilibrium — Graph Interpretation, Stresses & Kc
COCl₂(g) ⇌ CO(g) + Cl₂(g) ΔH = +107,6 kJ·mol⁻¹ (Kc at 395°C = 1,2 × 10³)
Concentration vs Time — COCl₂(g) ⇌ CO(g) + Cl₂(g)
t=5 min: Cl₂ added → [Cl₂] jumps → reverse reaction favoured → [CO] and [Cl₂] decrease, [COCl₂] increases until new equilibrium at t=10 min.
t=15 min: All concentrations drop simultaneously → volume increased → pressure decreased → forward reaction favoured (more moles of gas on product side) → new equilibrium at t=20 min.
t=25 min: Unknown stress — identify by calculating Kc from the new equilibrium concentrations and comparing to original Kc.
Explain what dynamic chemical equilibrium refers to. (2)
Identify the stress at t=5 min. State Le Chatelier's principle. (1+2)
Use Le Chatelier to explain changes between 5 and 10 minutes. (3)
Identify and explain the stress at t=15 min (all concentrations drop). (3)
Which of the rate vs time graphs below (A–D) is correct for the period 10–25 min? Solid line = forward reaction; dashed = reverse. (2)
Rate vs Time — 4 options (stress at t=15 min)
At t=15 min, pressure decreases. This favours the side with more moles of gas — the product side (2 mol vs 1 mol). Both rates increase instantly, but the forward rate increases more. They converge again at a new, higher equilibrium rate. Answer: B.
A is wrong — equal jumps mean no side is favoured. C is wrong — they must re-equalise at equilibrium. D is wrong — fwd must be higher than rev, not the reverse.
At t=25 min: [COCl₂]=0,009; [CO]=0,40; [Cl₂]=0,52 mol·dm⁻³. Calculate Kc. Using this, identify the stress at t=25 min and explain fully. (4+5)
Model Answers
a
Dynamic chemical equilibrium: a reversible reaction in which the forward and reverse reactions occur at the same rate, so concentrations of reactants and products remain constant.
b
Stress at t=5: Cl₂ was added (increased [Cl₂]). Le Chatelier: when an external stress is applied to a system at equilibrium, the system adjusts to counteract the stress.
c
↑[Cl₂] → system counteracts by decreasing [Cl₂] → reverse reaction favoured → [CO] and [Cl₂] decrease, [COCl₂] increases, until new equilibrium is reached.
d
All concentrations drop simultaneously → volume of the container increased → pressure decreased. Le Chatelier: system favours the side with more moles of gas (right side: 2 mol vs 1 mol) → forward reaction favoured.
e
Graph B: after a stress that favours the forward reaction (pressure decrease), both forward and reverse rates increase, but forward increases more — then both settle at a new equal rate.
f
Kc = [CO][Cl₂]/[COCl₂] = (0,40)(0,52)/(0,009) = 0,208/0,009 = 23,1. Original Kc = 1,2 × 10³. New Kc = 23,1 (much smaller) → more reactants, fewer products → reverse reaction was favoured → since the forward reaction is endothermic, a reverse-favouring change means temperature decreased. Kc decreased → stress was a decrease in temperature.
⚠ Examiner Tip To identify a temperature stress from Kc: if Kc decreases from the original, the reverse reaction was favoured. Ask yourself: which temperature change favours the reverse reaction? If reverse is exothermic → temperature increased. If reverse is endothermic → temperature decreased.
Equilibrium
[20]
Haber Process — Industrial Compromise, Ostwald Process Kc & Open/Closed Systems
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) operated at 450°C and 200 atm. A graph shows % NH₃ yield at different temperatures and pressures.
Haber Process — % NH₃ Yield vs Temperature at Different Pressures
— 400 atm
— 200 atm
— 100 atm
- - 50 atm
★ Industrial operating point (~16% yield at 450°C, 200 atm)
All four curves slope downward as temperature increases. By Le Chatelier's principle, increasing temperature favours the reverse (endothermic) reaction — so the reverse must absorb heat — meaning the forward reaction is exothermic. Higher pressure (400 atm) gives higher yield at all temperatures because the forward reaction produces fewer moles of gas (4 mol → 2 mol).
Ostwald process first step: 4NH₃(g) + 5O₂(g) ⇌ 4NO(g) + 6H₂O(g) in a 2 dm³ container. Kc between t=0 and t=5 min = 1,00 × 10⁻⁵.
Is the Haber forward reaction exothermic or endothermic? Explain using the graph and Le Chatelier. (1+4)
What is the % yield at 450°C, 200 atm? Will high pressure favour NH₃? (1+1)
Although 0°C gives >90% yield, industry uses much higher temperatures. Fully explain why. (3)
A catalyst is used. What effect does it have on % yield? (2)
Can Kc measured at 472°C and 300 atm be used for 450°C and 200 atm? Explain. (3)
Ostwald: if equilibrium [NO] = 1,7 mol·dm⁻³, use the ICE table below to calculate Kc. (7)
ICE Table — Ostwald Process: 4NH₃ + 5O₂ ⇌ 4NO + 6H₂O (2 dm³ container)
| NH₃ | O₂ | NO | H₂O | |
|---|---|---|---|---|
| Initial (mol) | 6,00 | 5,00 | 3,00 | 1,00 |
| Change (mol) | −0,40 | −0,50 | +0,40 | +0,60 |
| Equil (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) = (1,70)⁴(0,80)⁶ / ((2,80)⁴(2,25)⁵) = 6,18 × 10⁻⁴
Define open system. Removing NH₃ continuously — explain how this affects the yield with reference to reaction rate. (2+3)
Model Answers
a
Exothermic. From the graph, increasing temperature decreases the % yield of NH₃. By Le Chatelier, ↑T favours the reverse (endothermic) reaction — so the forward reaction must be exothermic.
b
~16% yield at 450°C, 200 atm. Yes, high pressure favours NH₃ (4 mol gas → 2 mol gas; system reduces pressure by producing fewer moles).
c
At 0°C the rate of reaction is extremely slow → NH₃ produced per day (total production) is very low. At 450°C, although equilibrium yield is lower, the reaction is fast enough to produce more NH₃ per unit time — the compromise between yield and rate maximises economic output.
d
No effect on % yield (equilibrium position). A catalyst speeds up both forward and reverse reactions equally → equilibrium reached faster, but same equilibrium position (same Kc).
e
No. Kc is temperature-dependent. The temperature used to measure it (472°C) is not the same as the industrial temperature (450°C). Kc must be measured at the same temperature to be valid.
f
ICE table (in 2 dm³): Initial mol: NH₃=6, O₂=5, NO=3, H₂O=1. Equilibrium NO=3,40 mol → Δ(NO)=+0,40 mol. Change: NH₃=−0,40, O₂=−0,50, NO=+0,40, H₂O=+0,60. Equilibrium: NH₃=5,60, O₂=4,50, NO=3,40, H₂O=1,60. Convert to mol·dm⁻³ (divide by 2). Kc = [NO]⁴[H₂O]⁶/([NH₃]⁴[O₂]⁵) = (1,70)⁴(0,80)⁶/((2,80)⁴(2,25)⁵) = 6,18 × 10⁻⁴. Since Kc increased (was 1×10⁻⁵, now 6,18×10⁻⁴) → forward endothermic reaction was favoured → temperature increased.
g
Open system: one in which both energy and matter can be exchanged with surroundings. Removing NH₃ continuously decreases the reverse reaction rate → forward reaction is favoured (more NH₃ produced) → favours yield of NH₃.
⚠ Examiner Tip "No effect" is a valid and often correct answer for catalysts on equilibrium yield. A catalyst cannot change where equilibrium lies — it can only change how quickly equilibrium is reached. Writing that a catalyst increases yield is always wrong.
Electrochemistry
[21]
Zn–Ag Galvanic Cell — Half-Reactions, Salt Bridge & Ecell
A Zn–Ag electrochemical cell operates under standard conditions. E°(Zn²⁺/Zn) = −0,76 V; E°(Ag⁺/Ag) = +0,80 V.
Zn–Ag Galvanic Cell
Zinc has a more negative reduction potential (−0,76 V) than silver (+0,80 V), so zinc is the anode (oxidised) and silver is the cathode (reduced). Electrons flow through the external wire from Zn to Ag. The salt bridge allows K⁺ and NO₃⁻ ions to flow between half-cells, maintaining electrical neutrality. E°cell = +0,80 − (−0,76) = +1,56 V.
Is this a galvanic or electrolytic cell? Explain. (3)
Write the reduction half-reaction (with phases). (2)
Write the oxidation half-reaction (with phases). (2)
Write the overall balanced ionic cell reaction. (2)
Lufuno says "the silver electrode will dissolve." Is she correct? Explain. (1+2)
Liam says "Zn²⁺ ions will move through the salt bridge and coat the silver electrode with zinc." Criticise this statement fully. (4)
Name 2 functions of the salt bridge. Calculate the standard emf. (2+3)
Model Answers
a
Galvanic (voltaic) cell — a spontaneous redox reaction occurs in each half-cell, generating an emf measured on the voltmeter.
b
Ag⁺(aq) + e⁻ → Ag(s)c
Zn(s) → Zn²⁺(aq) + 2e⁻d
Zn(s) + 2Ag⁺(aq) → Zn²⁺(aq) + 2Ag(s)e
No. The zinc electrode dissolves (is oxidised). The silver electrode gains mass as Ag⁺ ions are reduced to Ag metal at the cathode.
f
Liam is partially wrong. Ag⁺ ions in the cathode electrolyte are reduced to Ag(s) which coats the Ag electrode — correct. However, Zn²⁺ ions do NOT move through the salt bridge to coat the silver electrode. Zn²⁺ moves through the salt bridge only to replace the Ag⁺ ions consumed at the cathode and restore electrical neutrality. The coating is Ag, not Zn.
g
Salt bridge: (1) completes the circuit; (2) maintains electrical neutrality in the electrolyte solutions. E°cell = E°cathode − E°anode = 0,80 − (−0,76) = +1,56 V
⚠ Examiner Tip Salt bridge "criticise" questions carry 4 marks — you need 4 distinct, correct points. Cover: what actually happens at the cathode, why the ion concentration in the cathode half-cell changes, what the salt bridge ions actually do, and why the learner's stated outcome is wrong.
Electrochemistry
[26]
Standard Hydrogen Electrode, Cell Notation & Designing a Cell for Target Ecell
The SHE is connected to unknown metal M. Voltmeter reads +0,80 V. Substances available: Fe, FeSO₄, Fe(NO₃)₃, Ag, AgNO₃, KNO₃.
Standard Hydrogen Electrode (SHE) Connected to Unknown Metal M
The voltmeter reads +0,80 V. The SHE is always the reference (E° = 0,00 V) and is the anode here. Metal M is the cathode with E° = +0,80 V — this identifies M as Silver (Ag). Standard conditions: all aqueous ions at 1 mol·dm⁻³, gas pressure 1 atm, temperature 25°C. The salt bridge must contain KNO₃ (never a halide — AgCl would precipitate).
Define an anode. State the conditions for standard electrode potentials. (2+3)
From the +0,80 V reading: identify metal M; label anode/cathode; write the cell notation including phase labels. (4)
State the functions of the salt bridge. (2)
If [HNO₃] is increased in the SHE half-cell, will the voltmeter reading increase, decrease or stay the same? Fully explain. (3)
Using the substances provided, identify the two half-reactions needed to produce a cell with E°cell = +0,84 V. Include an Ecell calculation to support your answer. Write the balanced overall ionic reaction. (6+2)
Model Answers
a
Anode: the electrode where oxidation occurs. Standard conditions: aqueous ion concentration = 1 mol·dm⁻³; gas pressure = 1 atm; temperature = 25°C.
b
M is Silver (Ag), E° = +0,80 V. Ag is the cathode (reduction). SHE is the anode. Cell notation: Pt(s) | H₂(g) | H⁺(aq) || Ag⁺(aq) | Ag(s). Salt bridge electrolyte: KNO₃ (cannot use a halide salt).
c
(1) Completes the circuit; (2) maintains electrical neutrality in both half-cells.
d
Decreases. Increasing [H⁺] is a product increase in the cathode half-cell, favouring the reverse reaction (H⁺ being produced). This decreases the driving force and reduces the emf.
e
Oxidation: Fe → Fe³⁺ + 3e⁻ (E° = −(−0,04) = +0,04 V as anode). Reduction: Ag⁺ + e⁻ → Ag (E° = +0,80 V). E°cell = +0,80 − (−0,04) = +0,84 V. Overall: Fe + 3Ag⁺ → Fe³⁺ + 3Ag.
⚠ Examiner Tip When designing a cell to meet a target Ecell, use trial-and-error with the reduction potential table: E°cell = E°cathode − E°anode. The cathode must always have the higher (more positive) reduction potential. Also note: KNO₃ in the salt bridge — never use a halide salt (KCl, KBr) when Ag⁺ is present because AgCl precipitate forms.
Electrochemistry
[25]
Sn–Unknown Metal X Cell — Identifying Metal, Faraday & Cell Notation
Cell: Sn electrode (mass increases when switch closed) and unknown Metal X. E°(Sn²⁺/Sn) = −0,14 V. E°cell = +0,60 V.
Name the energy conversion; identify which electrode (Sn or X) is the cathode; define anode. (2+2+2)
State the direction of electron flow in the external circuit. (1)
Calculate E° for the X half-cell; hence identify metal X. (3+1)
Write the cell notation (standard conditions, phase labels). Write the balanced net ionic equation. (4+3)
Suggest a suitable salt bridge electrolyte. Explain how the salt bridge maintains neutrality at the cathode. (1+3)
The salt bridge is replaced by one that is narrower and longer. State how this affects: emf; internal resistance; current delivered. (3)
Model Answers
a
Chemical → electrical. Cathode: Sn (gains mass → Sn²⁺ ions are being reduced to Sn metal at this electrode). Anode: electrode where oxidation occurs.
b
From X to Sn (electrons flow from anode to cathode in external circuit).
c
E°cell = E°cathode − E°anode. +0,60 = −0,14 − E°(X). E°(X) = −0,14 − 0,60 = −0,74 V. Metal X = Chromium (Cr).
d
Cell notation: X(s)/X³⁺(aq)(1 mol·dm⁻³) // Sn²⁺(aq)(1 mol·dm⁻³)/Sn(s) @ 25°C. Net ionic: 2X(s) + 3Sn²⁺(aq) → 2X³⁺(aq) + 3Sn(s).
e
Salt bridge: sodium/potassium nitrate (KNO₃). At the cathode, Sn²⁺ is being reduced (consumed) → the electrolyte solution becomes more negative. Positive ions from the salt bridge migrate into the cathode solution to restore neutrality; negative ions migrate out into the salt bridge.
f
Emf: remains the same (determined by electrode potentials only). Internal resistance: increases (narrower, longer path = more resistance). Current delivered: decreases (I = V/R; same emf, higher resistance).
⚠ Examiner Tip The emf (Ecell) of a galvanic cell is determined only by the electrode potentials — not by the shape of the salt bridge or the distance between half-cells. Many learners incorrectly state that narrowing/lengthening the salt bridge reduces the emf. It doesn't. It reduces current by increasing resistance.
Electrochemistry
[14]
Electrolytic Cells — CuCl₂ Electrolysis, Hall-Héroult Cell & Faraday Calculations
CuCl₂(aq) → Cu(s) + Cl₂(g). Two graphite electrodes in CuCl₂ solution. 0,011 mol Cu forms in 10 minutes.
Electrolytic Cell — Electrolysis of CuCl₂(aq)
Cathode (−): Cu²⁺(aq) + 2e⁻ → Cu(s) — copper metal deposits on the electrode.
Anode (+): 2Cl⁻(aq) → Cl₂(g) + 2e⁻ — chlorine gas is released.
To ensure Cl₂ (not O₂) is the main gas at the anode, the solution must have a high concentration of Cl⁻ ions. The reduction potentials for Cl₂ (+1,36 V) and O₂ (+1,23 V) are similar, but at high [Cl⁻] the rate of Cl⁻ oxidation dominates.
At which electrode (anode or cathode) will Cl₂ form? (1)
What must the student do to ensure Cl₂ (not O₂) is the main gas at that electrode? Explain with reference to half-reactions. (3)
Calculate the current supplied to form 0,011 mol Cu in 10 minutes. (6)
Hall-Héroult cell for Al extraction: give the formula for aluminium oxide; write the cathode half-reaction; state the purpose of cryolite. (1+2+1)
Model Answers
a
Anode (oxidation of Cl⁻ ions: 2Cl⁻ → Cl₂ + 2e⁻).
b
Use a high concentration of CuCl₂. The electrode potentials for Cl₂ (+1,36 V) and O₂ (+1,23 V) are very similar. H₂O is inherently more reactive than Cl⁻. However, at high [Cl⁻] concentration, the oxidation of Cl⁻ predominates (at a greater rate than H₂O oxidation), so Cl₂ is the main product.
c
Cu²⁺ + 2e⁻ → Cu, so n(Cu):n(e⁻) = 1:2. n(e⁻) = 2 × 0,011 = 0,022 mol. Q = n × F = 0,022 × 96500 = 2123 C. t = 10 min = 600 s. I = Q/t = 2123/600 = 3,54 A
d
Formula: Al₂O₃. Cathode: Al³⁺ + 3e⁻ → Al. Cryolite: reduces the melting point of aluminium oxide to ~950°C (from ~2000°C), reducing energy costs.
⚠ Examiner Tip Faraday calculation sequence: n(product) → n(electrons) using mole ratio → Q = nF → I = Q/t. Always convert time to seconds. The most common error is forgetting to multiply n(product) by the number of electrons in the half-reaction (2 for Cu²⁺, 3 for Al³⁺).
Reaction Rates
[33]
CaCO₃ + HCl — Rate Calculations, Graphs & Collision Theory
CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + CO₂(g) + H₂O(l) ΔH < 0
Define reaction rate in terms of this investigation. (2)
Explain using collision theory why Experiment 2 (higher concentration HCl) has a higher rate than Experiment 1. (3)
Calculate the volume of HCl (in cm³) that reacts with 4 g CaCO₃ if HCl is the limiting reagent (concentration 0,4 mol·dm⁻³). (5)
Sketch the potential energy diagram for this reaction (ΔH < 0). Label Ea, activated complex, and heat of reaction. (4)
Potential Energy Diagram — CaCO₃ + HCl (Exothermic, ΔH < 0)
(a) Heat of reaction (ΔH): The energy difference between reactants and products. Negative here — products are at lower energy than reactants → exothermic.
(b) Activation energy (Ea): The minimum energy particles must have for a successful collision. Shown as the energy difference between reactants and the peak.
(c) Activated complex: The unstable, high-energy transition state at the peak of the curve. It is not a product — it breaks down to give either reactants or products.
Using the mass-loss data table below, calculate the average rate between t=10 s and t=30 s in mol·s⁻¹. (4)
Mass-Loss Data — CaCO₃ + HCl Reaction at 20°C
| Time (s) | 0 | 15 | 30 | 45 | 60 | 75 | 90 | 105 | 120+ |
|---|---|---|---|---|---|---|---|---|---|
| Mass (g) | 170,0 | 169,7 | 169,5 | 169,4 | 169,3 | 169,2 | 169,1 | 169,1 | 169,0 |
Rate = Δmass/Δt → convert to mol·s⁻¹ by dividing by M(CO₂) = 44 g·mol⁻¹
State and explain what happens to the rate when large lumps replace small granules (using collision theory). (4)
Calculate the percentage purity of a 1,5 g impure CaCO₃ sample that produces 180 cm³ of CO₂ at STP with excess HCl. (6)
Model Answers — Key Steps
a
Change in concentration (or volume) of CO₂ produced (or HCl consumed) per unit time.
b
Higher concentration = more moles per unit volume → more collisions per unit time → more effective collisions per unit time → higher reaction rate.
c
n(CaCO₃) = 4/100 = 0,04 mol | ratio CaCO₃:HCl = 1:2 | n(HCl) = 0,08 mol | V = n/c = 0,08/0,4 = 0,2 dm³ = 200 cm³
d
Diagram: exothermic shape (products lower than reactants). Label Ea above reactants peak, activated complex at peak, ΔH = heat of reaction (vertical drop from reactants to products).
e
From graph: Δmass from t=10 to t=30 = 164,85 − 164,73 = 0,12 g (read from graph/table). Average rate = −0,12/20 = −0,006 g·s⁻¹. In mol: 0,006/44 = 1,36 × 10⁻⁴ mol·s⁻¹
f
Rate decreases. Large lumps have smaller surface area → fewer reactant particles exposed per unit time → fewer collisions per unit time → fewer effective collisions per unit time → lower rate.
g
n(CO₂) = V/Vm = 0,180/22,4 = 0,00804 mol | n(CaCO₃) = 0,00804 mol | m(CaCO₃) = 0,00804 × 100 = 0,804 g | % purity = 0,804/1,5 × 100 = 53,6%
⚠ Examiner Tip Rate in mol·s⁻¹ from mass-loss data: divide the mass rate (g·s⁻¹) by the molar mass of the GAS (CO₂ = 44 g·mol⁻¹). A very common error is dividing by the molar mass of CaCO₃ instead.
Quantitative Chemistry
[25]
Chalcopyrite — Moles, Stoichiometry & Limiting Reagent
Reaction 1: 2CuFeS₂(s) + 3O₂(g) → 2FeO(s) + 2CuS(s) + 2SO₂(g)
77,8 g of chalcopyrite (CuFeS₂) is reacted with oxygen.
Reaction 2: CuS(s) + O₂(g) → Cu(s) + SO₂(g)
CuS from Reaction 1 is reacted with 26 g of O₂ in a separate vessel.
Calculate the number of moles in 77,8 g of chalcopyrite (M = 183,5 g·mol⁻¹). (3)
Use Reaction 1 to calculate the mass of CuS formed. (4)
Determine and fully explain which reactant is limiting in Reaction 2. (4)
What volume of SO₂ is produced in Reaction 2 (at STP)? (3)
Model Answers
a
n = m/M = 77,8/183,5 = 0,42 mol
b
Ratio CuFeS₂ : CuS = 2:2 = 1:1 | n(CuS) = 0,42 mol | M(CuS) = 95,5 g·mol⁻¹ | m = 0,42 × 95,5 = 40,11 g
c
n(O₂) = 26/32 = 0,81 mol | Ratio CuS:O₂ = 1:1 | Need 0,42 mol O₂ for 0,42 mol CuS. Have 0,81 mol O₂ — more than enough. ∴ CuS is limiting; O₂ is in excess.
d
Ratio CuS:SO₂ = 1:1 | n(SO₂) = 0,42 mol | V = nVm = 0,42 × 22,4 = 9,41 dm³
⚠ Examiner Tip For limiting reagent proofs: don't just state which is limiting — calculate moles of BOTH reactants, compare to the required mole ratio, and explicitly state "X is limiting because we have less than needed / Y is in excess." Incomplete proofs lose marks.
Bonding & IMF
[15]
HCl Polarity, Aluminium Oxide & Intermolecular Forces
Define covalent bond. (2)
Define electronegativity. (2)
With reference to the difference in electronegativity between H and Cl, explain why the bond in HCl is polar. (3)
Aluminium oxide has a very high melting point. With reference to the forces holding its particles together, explain why. (4)
Define intermolecular force. (2)
Compound C has significantly higher viscosity than compound B. Account for this difference with reference to the relevant intermolecular forces in each. (4)
Model Answers
a
A covalent bond is the sharing of a pair of electrons between two atoms (when atomic orbitals overlap).
b
Electronegativity is the measure of the tendency of an atom in a molecule to attract the shared (bonding) electrons towards itself.
c
Chlorine has a higher electronegativity than hydrogen. The shared electrons are pulled closer to Cl → Cl becomes δ⁻ and H becomes δ⁺. This unequal sharing of electrons creates a permanent dipole → polar bond.
d
Al₂O₃ is an ionic compound. It consists of Al³⁺ and O²⁻ ions in a giant ionic lattice. The electrostatic forces of attraction between oppositely charged ions are very strong and operate in all directions. Much energy is required to overcome these strong ionic bonds → very high melting point.
e
An intermolecular force is a force of attraction between molecules (not within molecules).
f
Compound C has stronger IMFs than Compound B (e.g., hydrogen bonding vs London forces). Molecules of C are more strongly attracted to each other → greater resistance to flow → higher viscosity.
⚠ Examiner Tip "Intramolecular" = within a molecule (bonds). "Intermolecular" = between molecules (forces). These terms are frequently confused in exam answers — a wrong term = lost marks.
Bonding & IMF
[22]
HF vs F₂, SiO₂ Structure & KBr Solubility
Define ionic bond. (2)
Name the specific intramolecular bond in HF; in F₂. (2+2)
Name the intermolecular forces between HF molecules; between F₂ molecules. (2+2)
Explain why HF (bp 19,5°C) has a higher boiling point than F₂ (bp −188°C). (2)
SiO₂ has a very high melting point of 1610°C. With reference to structure and bonds, explain why. (4)
KBr is very soluble in water but insoluble in cyclohexane (C₆H₁₂). Explain why KBr cannot dissolve in C₆H₁₂. (2)
Model Answers
a
An ionic bond is the electrostatic force of attraction between oppositely charged ions.
b
HF: polar covalent bond | F₂: (pure/non-polar) covalent bond
c
HF molecules: hydrogen bonds | F₂ molecules: London forces (dispersion forces)
d
HF molecules form hydrogen bonds (stronger IMFs) while F₂ molecules only have London forces (weaker). More energy is required to overcome hydrogen bonds → higher boiling point for HF.
e
SiO₂ has a giant covalent (network) structure. Each Si atom is covalently bonded to 4 oxygen atoms forming a continuous 3D network. These covalent bonds extend throughout the entire structure and are very strong. Much energy is needed to break all the covalent bonds → very high melting point.
f
C₆H₁₂ is a non-polar solvent with only London forces. KBr is an ionic compound with strong ionic bonds. The weak London forces of C₆H₁₂ cannot overcome the strong ionic forces in KBr to dissociate the ions → KBr cannot dissolve ("like dissolves like").
⚠ Examiner Tip For solubility questions: always state why the solvent CANNOT provide enough energy to overcome the solute's IMFs. Just saying "like dissolves like" alone is not enough for full marks — you must reference the specific forces involved.
Definition Bank
Examinable definitions across all P2 topics — learn the exact wording.
⚗️ Organic Chemistry
Homologous series
A series of compounds with the same functional group and general formula, in which each member differs from the previous by a CH₂ unit.
Functional group
An atom or group of atoms that forms the centre of chemical activity in a molecule.
Structural isomers
Compounds with the same molecular formula but different structural formulae.
Chain isomers
Structural isomers with the same molecular formula but different carbon chain arrangements (e.g. pentane and methylbutane).
Functional isomers
Structural isomers with the same molecular formula but different functional groups (e.g. an ester and a carboxylic acid).
Positional isomers
Structural isomers with the same molecular formula and functional group, but the functional group is in a different position on the carbon chain.
Saturated compound
An organic compound in which all carbon-carbon bonds are single bonds (no double or triple bonds).
Unsaturated compound
An organic compound containing one or more carbon-carbon double or triple bonds.
Hydrocarbon
An organic compound containing only carbon and hydrogen atoms.
🔗 Bonding & IMF
Covalent bond
The sharing of a pair of electrons between two atoms (formed when atomic orbitals overlap).
Ionic bond
The electrostatic force of attraction between oppositely charged ions.
Electronegativity
The measure of the tendency of an atom in a molecule to attract the bonding (shared) electrons towards itself.
Polar covalent bond
A covalent bond in which the bonding electrons are unequally shared due to a difference in electronegativity between the two atoms.
Intermolecular force
A force of attraction between molecules (not within molecules).
Intramolecular bond
A force of attraction within a molecule, between atoms in the same molecule (e.g. covalent bond, ionic bond).
London forces
Weak, temporary IMFs that arise due to temporary dipoles caused by uneven distribution of electrons. Present in all molecules.
Hydrogen bond
A strong intermolecular force between a hydrogen atom (bonded to N, O, or F) and a lone pair of electrons on a neighbouring N, O, or F atom.
⚖️ Acids & Bases
Arrhenius acid
A substance that produces H⁺ (or H₃O⁺) ions when dissolved in water.
Arrhenius base
A substance that produces OH⁻ ions when dissolved in water.
Brønsted-Lowry acid
A proton (H⁺) donor.
Brønsted-Lowry base
A proton (H⁺) acceptor.
Conjugate acid-base pair
Two species that differ by a single proton (H⁺) — the acid donates the proton to form its conjugate base.
Amphoteric substance
A substance that can act as both an acid and a base (e.g. water, HSO₄⁻).
Standard solution
A solution of accurately known concentration.
Concentration
The amount of solute (in mol) per unit volume of solution (in dm³). Units: mol·dm⁻³.
🔄 Chemical Equilibrium
Chemical equilibrium
The state in a reversible reaction where the rate of the forward reaction equals the rate of the reverse reaction and the concentrations of reactants and products remain constant.
Le Chatelier's principle
When the equilibrium of a closed system is disturbed, the system will re-establish equilibrium by favouring the reaction that opposes the disturbance.
Equilibrium constant (Kc)
The ratio of the concentrations of products to reactants at equilibrium, with each concentration raised to the power of its stoichiometric coefficient.
⚡ Electrochemistry
Galvanic cell
An electrochemical cell in which a spontaneous redox reaction converts chemical energy to electrical energy.
Electrolytic cell
An electrochemical cell in which electrical energy is used to drive a non-spontaneous redox reaction.
Standard electrode potential
The potential (voltage) of a half-cell measured relative to the standard hydrogen electrode under standard conditions (25°C, 1 mol·dm⁻³, 101,3 kPa).
Anode
The electrode where oxidation occurs. In a galvanic cell, the negative electrode.
Cathode
The electrode where reduction occurs. In a galvanic cell, the positive electrode.
💥 Reaction Rates & 🧮 Quantitative Chemistry
Reaction rate
The change in concentration of a reactant or product per unit time.
Activation energy (Ea)
The minimum energy required by colliding particles for a reaction to occur.
Catalyst
A substance that increases the rate of a reaction by providing an alternative pathway with a lower activation energy, without being consumed in the reaction.
Limiting reagent
The reactant that is completely consumed in a reaction and determines the maximum amount of product that can be formed.
Percentage yield
(Actual yield ÷ Theoretical yield) × 100%. The ratio of the actual amount of product obtained to the theoretical maximum.
Percentage purity
(Mass of pure substance ÷ Total mass of sample) × 100%.
Molar volume at STP
22,4 dm³·mol⁻¹ — the volume occupied by 1 mole of any gas at standard temperature and pressure (0°C, 101,3 kPa).
Exam Strategy
Proven approaches for maximising marks in P2 Chemistry.
1
Time ManagementP2 is 150 marks in 3 hours = 1,2 minutes per mark. Use this as your benchmark.
- MCQ (Q1, ~20 marks): spend max 15–18 min. Never spend more than 90 sec on one MCQ.
- Organic Q (≈35 marks): allocate 42 min. Reaction type + name questions are fast marks.
- Calculations: show every step — partial marks are awarded even if the final answer is wrong.
- Leave 10 min at the end to re-read all your "define" answers.
2
Organic Chemistry — Fast Marks- Reaction type questions: always give the specific type, not just general. "Halogenation" not just "substitution".
- IUPAC naming: count the longest chain first. Number from the end closest to the functional group.
- Boiling point explanations: 4-step structure — (1) identify IMF type, (2) state which is stronger and why, (3) more energy required to overcome, (4) therefore higher boiling point.
- Ester questions: always mention both conditions — water bath (flammable reactants) AND concentrated H₂SO₄ catalyst.
- Combustion equations: balance O₂ last. Check H₂O is correct before finalising.
3
Calculations — Never Lose Method Marks- Always write the formula first:
n = m/M,c = n/V,V = nVm - Substitute values with units:
n = 77,8/183,5 = 0,42 mol - Show mole ratio explicitly: "ratio CuFeS₂ : CuS = 2:2 = 1:1"
- For limiting reagent: calculate moles of BOTH, compare to ratio, conclude explicitly.
- Round only at the final answer (not intermediate steps).
4
Definition Questions — Every Mark CountsDefinitions are guaranteed marks — learn the exact IEB wording for these:
- Homologous series, functional group, structural isomer — always asked
- Reaction rate, activation energy, limiting reagent — always asked
- Equilibrium, Le Chatelier, Kc — always asked
- Galvanic cell, standard electrode potential — always asked
- Use the Definition Bank tab to drill these before every test.
5
Common Organic Errors to Avoid- Calling a compound with –OH a "hydrocarbon" (it's not — hydrocarbons contain only C and H)
- Giving "addition" when asked for specific type of addition (hydrogenation? hydration? hydrohalogenation?)
- Drawing structural formulae for a "condensed structural formula" question
- Using conc H₂SO₄ in water for elimination — this gives substitution. Use conc KOH in ethanol for elimination.
- Forgetting to discard outlier titration results when calculating average volume
Formula Trainer
Flashcard drill for all P2 formulas. Mark each as "Know it" or "Review" to track progress.
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