The S6 Chemistry II 2024 National Examination, sat on 26 July 2024 under NESA, covers atomic structure, acid-base theory, redox chemistry, organic chemistry, coordination compounds, chemical kinetics, and thermodynamics. This paper is set for the BCG, MCB, PCB, PCM, and ANP combinations. Below are fully worked solutions to every question, with formulas and step-by-step calculations shown throughout.
Since this paper spans several combinations, you may also find our other subject worked solutions useful depending on your own combination: our S6 Biology II worked solutions pair naturally with BCG, MCB, and PCB, our S6 Physics II 2024 worked solutions and S6 Physics 2025 worked solutions cover the PCB and PCM combinations, our S6 Mathematics II 2024 and 2025 worked solutions support MCB and PCM, and our S6 Geography II worked solutions complete the BCG combination. Every past-paper article we publish is also organized on the NESA past papers worked solutions hub.
Paper Overview: S6 Chemistry II 014
The paper has two sections. Section A is compulsory and worth 70 marks, covering atomic structure, acid-base theory, redox reactions, organic compounds, bonding, coordination chemistry, and equilibrium. Section B requires candidates to attempt only 3 of 5 longer questions worth 30 marks, covering kinetics, periodic trends in bonding, polymers, Born-Haber cycles, and titration curves. Students do not need the periodic table for this paper, and silent, non-programmable calculators are permitted, with a pencil required for drawing. You have 3 hours to complete the paper.
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Section A: Attempt All Questions (70 marks)
Question 1: Key Terms in Atomic Structure
a) Atoms of the same element with the same atomic number but different mass numbers: Isotopes.
b) A device that separates gaseous ions by mass-to-charge ratio and records the resulting spectrum: Mass spectrometer.
c) The term indicating an element's position in the periodic table, or the number of protons in the nucleus: Atomic number (Z).
d) A sub-atomic particle that is electrically neutral: Neutron.
Question 2: True or False
a) The arrangement of atoms in graphite structure is called hexagonal. Answer: True. Graphite consists of layers of carbon atoms arranged in hexagonal rings, bonded covalently within each layer.
b) Considering the stability of group 14 elements, lead (II) oxide is more stable than lead (IV) oxide. Answer: True. This reflects the inert pair effect, where heavier group 14 elements like lead increasingly favor the +2 oxidation state over +4 as you descend the group, making PbO more thermodynamically stable than PbO₂.
c) In diamond structure, there are delocalized electrons. Answer: False. Diamond has all four valence electrons of each carbon atom localized in strong, directional covalent bonds in a rigid tetrahedral lattice, with no delocalized electrons, which is why diamond does not conduct electricity, unlike graphite.
d) Lead (IV) chloride decomposes readily without heating to form PbCl₂ and Cl₂. Answer: True. This is again due to the inert pair effect: Pb(IV) compounds are relatively unstable and tend to decompose to the more stable Pb(II) state, releasing chlorine gas.
Question 3: Acid-Base Theory
a) According to Arrhenius' theory, a base is defined as: Answer: iv) Any substance that dissociates in aqueous solution to produce hydroxyl ions (OH⁻) as the only negative ions. This is the defining feature of the Arrhenius definition, which is narrower than the Brønsted-Lowry or Lewis definitions since it only applies to aqueous solutions.
b) In Cu²⁺ + 4NH₃ ⇌ [Cu(NH₃)₄]²⁺, Cu²⁺ ions are acting as: Answer: iii) A Lewis acid. A Lewis acid is an electron-pair acceptor. Here, the Cu²⁺ ion accepts lone pairs of electrons from the nitrogen atoms in ammonia to form coordinate (dative) bonds, which is the hallmark of Lewis acid-base chemistry, distinct from proton transfer reactions.
c) Which species is more likely to act as amphiprotic (amphoteric)? Answer: i) HS⁻. HS⁻ (hydrogen sulfide ion) can both donate a proton (to form S²⁻) and accept a proton (to reform H₂S), making it amphiprotic. N₂ has no acidic hydrogen to donate or accept protons with, S²⁻ can only accept a proton, and CN⁻ can only accept a proton.
d) The conjugate acid for phosphate ions (PO₄³⁻) is: Answer: ii) HPO₄²⁻. A conjugate acid is formed by adding one proton (H⁺) to a base. Adding one H⁺ to PO₄³⁻ gives HPO₄²⁻, reducing the overall negative charge by one.
Question 4: Redox Vocabulary Fill in the Blanks
a) An oxidizing agent is a substance or chemical species containing the element or species that accepts electrons, allowing another element or species to be oxidized.
b) Disproportionation is a redox reaction where an element in a molecule or chemical species is simultaneously oxidized and reduced.
c) When balancing redox reactions, it is required to sum up the two half-reactions.
d) The oxidation state is the total number of electrons lost or gained by an atom of element in a chemical combination.
Question 5: Matching Alcohols to Their Uses
Step-by-step reasoning: Methanol is a well-known industrial feedstock for formaldehyde production. Ethanol is the alcohol found in alcoholic drinks and is widely used as a biofuel and solvent. Propan-2-ol (isopropanol) is the classic antiseptic alcohol used to clean skin and wounds. Propan-1-ol is used industrially as a solvent for resins and in making propanoic acid derivatives.
Answer:
A. Propan-1-ol → 4) Solvent for resins, lacquers; manufacture of propanoic acid; making plasticizers
B. Ethanol → 2) Used as biofuel; solvent for paints, varnishes; alcoholic drink
C. Methanol → 1) Manufacture of formaldehyde, dyes, plastics
D. Propan-2-ol → 3) Solvent for oils; synthesis of acetone; manufacture of antiseptic solutions
Question 6: Titration Concepts
a) The equivalence point occurs when: Answer: iv) The numbers of equivalents for the two solutions become equal. This is the defining condition of the equivalence point, regardless of what color change, if any, an indicator shows.
b) A specialized acid-base titration used to determine the concentration of a basic solution is known as: Answer: i) Acidimetry. Acidimetry uses a standard acid to determine the concentration of an unknown base, while alkalimetry uses a standard base to determine the concentration of an unknown acid.
c) The most suitable acid for acidifying a KMnO₄/Fe²⁺ titration is: Answer: iii) Sulphuric acid, since it is neither oxidized nor reduced during the reaction. Nitric acid is itself a strong oxidizing agent and would interfere by oxidizing Fe²⁺ independently, while hydrochloric acid can be oxidized by KMnO₄ to chlorine gas, introducing a side reaction. Sulphuric acid remains chemically inert to the redox process, making it the ideal choice.
d) A primary standard must have all of the following properties EXCEPT: Answer: iii) Insoluble in the solvent used and under the conditions used. A primary standard must actually dissolve readily and completely in the solvent to be usable in a titration; the other three properties (purity, availability, and stability in air) are genuine requirements.
Question 7: Nuclear Chemistry
a) Define radioactive isotope: A radioactive isotope is a version of an element whose nucleus is unstable and spontaneously emits radiation, such as alpha, beta, or gamma radiation, as it decays into a more stable nuclide.
b) One property of alpha particles: Alpha particles have a relatively large mass and a +2 charge, equivalent to a helium nucleus, which gives them low penetrating power, meaning they can be stopped by a sheet of paper or a few centimeters of air.
c) Deducing the decay chain for Th-232:
Step by step: An alpha particle is a helium nucleus (mass 4, atomic number 2), so alpha emission decreases the mass number by 4 and the atomic number by 2. A beta particle is an electron, so beta emission leaves the mass number unchanged but increases the atomic number by 1, since a neutron converts into a proton.
Starting from Th-232 (mass 232, atomic number 90):
After alpha emission → X: mass = 232 − 4 = 228 (b), atomic number = 90 − 2 = 88 (a). This makes X Radium-228.
After beta emission → Y: mass = 228 (d, unchanged), atomic number = 88 + 1 = 89 (c). This makes Y Actinium-228.
After beta emission → Z: mass = 228 (f, unchanged), atomic number = 89 + 1 = 90 (e). This makes Z Thorium-228.
Answer: a=88, b=228, c=89, d=228, e=90, f=228.
Question 8: Extraction of Copper from Chalcopyrite
a) i) Balanced equation for the second stage (copper(I) sulphide roasted with oxygen, giving copper(II) oxide and sulphur dioxide):
Cu₂S + 2O₂ → 2CuO + SO₂
ii) Balanced equation for the third stage (copper(II) oxide reduced by heating with carbon):
2CuO + C → 2Cu + CO₂
b) One use of copper: Copper is widely used in electrical wiring, since it is an excellent conductor of electricity and is relatively ductile and corrosion-resistant.
c) One environmental and one health-related problem from copper extraction: Environmental: the roasting stages release large quantities of sulphur dioxide gas, which contributes to acid rain and air pollution around smelting sites. Health: workers and nearby communities can suffer respiratory problems from inhaling sulphur dioxide and particulate matter released during smelting, and long-term exposure to copper dust and fumes can cause metal fume fever and other respiratory irritation.
Question 9: Electrolysis of Sodium Chloride and Copper Sulphate Solutions
a) Gas given off at the cathode, electrolysis of concentrated NaCl(aq): Hydrogen gas (H₂). In concentrated aqueous NaCl, water is preferentially reduced at the cathode over Na⁺ ions, since Na⁺ is very difficult to reduce in aqueous solution.
b) Product formed at the anode: Chlorine gas (Cl₂). In concentrated NaCl solution, Cl⁻ ions are preferentially oxidized at the anode over water, due to the high concentration of chloride ions.
c) Ionic equation for the anode reaction:
2Cl⁻(aq) → Cl₂(g) + 2e⁻
d) Replacing NaCl(aq) with CuSO₄(aq):
i) Product formed at the anode: Oxygen gas (O₂). With carbon (inert) electrodes and no halide ions present, water is oxidized at the anode instead, since SO₄²⁻ ions are very difficult to oxidize.
ii) Ionic equation for the cathode reaction: Since Cu²⁺ is easier to reduce than water, copper metal is deposited at the cathode:
Cu²⁺(aq) + 2e⁻ → Cu(s)
Question 10: Compounds M and N
Compound M shows a symmetric chain: CH₃-CO-CH₃, a ketone, propanone, and compound N shows CH₃-CH₂-CHO, an aldehyde, propanal.
a) Common molecular formula for M and N: Both have the formula C₃H₆O. This makes them structural isomers of each other, both containing a carbonyl group but positioned differently.
b) The relationship between M and N: Functional group isomerism (or more broadly, structural isomerism), since they share the same molecular formula but the carbonyl group occupies a different position, creating two different functional groups, ketone versus aldehyde.
c) Two further examples with four carbon atoms (C₄), one related to M, a ketone, and one related to N, an aldehyde: Related to M (ketone): butan-2-one (CH₃-CO-CH₂-CH₃). Related to N (aldehyde): butanal (CH₃-CH₂-CH₂-CHO).
d) What is observed when Brady's reagent (2,4-dinitrophenylhydrazine) is reacted with M and N: Both M and N contain a carbonyl group (C=O), so both react with Brady's reagent to form an orange or yellow precipitate, a 2,4-dinitrophenylhydrazone derivative, confirming the presence of an aldehyde or ketone functional group in each.
Question 11: Bonding in Acetonitrile (CH₃-C≡N)
a) Bond angle around the functional carbon (C₁, bonded to N): C₁ is bonded to only two groups overall, C₂ and N, joined by a triple bond to nitrogen, giving it a linear electron geometry. Answer: 180°.
b) Hybridizations on both carbon atoms: C₁, the nitrile carbon, forms one sigma bond to C₂ and a triple bond, one sigma plus two pi, to nitrogen, giving it two regions of electron density arranged linearly, so C₁ is sp hybridized. C₂, the methyl carbon, is bonded to three hydrogen atoms and to C₁, giving it four regions of electron density in a tetrahedral arrangement, so C₂ is sp³ hybridized.
c) Total sigma and pi bonds in acetonitrile: Counting each bond: 3 C–H bonds, each sigma, 1 C–C bond, sigma, and the C≡N triple bond, 1 sigma plus 2 pi. This gives 5 sigma bonds and 2 pi bonds in total.
Question 12: Coordination Compounds
a) Define coordination number: The coordination number of a complex compound is the total number of donor atoms, or ligand attachment points, directly bonded to the central metal atom or ion.
b) i) Name or formula: Na₂[CuBr₄] is named sodium tetrabromocuprate(II). The complex ion [CuBr₄]²⁻ has copper in the +2 oxidation state, since each Br⁻ contributes −1, four bromides give −4, balanced by two Na⁺ giving +2 overall for the complex ion, meaning Cu is +2.
ii) Name or formula: Triamminetriaquachromium(III) ion has the formula [Cr(NH₃)₃(H₂O)₃]³⁺. This follows directly from the name: three ammine (NH₃) ligands, three aqua (H₂O) ligands, and chromium in the +3 oxidation state.
c) Shape of the complex containing Pt(II) and four chloro ligands (square planar): [PtCl₄]²⁻ adopts a square planar geometry, with the four chloride ligands positioned at the four corners of a square around the central platinum ion, all in the same plane, with bond angles of 90° between adjacent ligands.
Question 13: Reaction Quotient and Equilibrium Position
The reaction: CH₄(g) + 2H₂S(g) ⇌ 4H₂(g) + CS₂(g), with Kc = 0.036 at 96°C. In a 250 mL vessel: 2.0 mol CH₄, 2.0 mol H₂S, 1.0 mol H₂, 1.0 mol CS₂.
a) Expression for Kc:
$$K_c = \frac{[H_2]^4[CS_2]}{[CH_4][H_2S]^2}$$b) Calculating Qc: First, convert moles to concentrations by dividing by the 0.250 L volume:
$$[CH_4] = \frac{2.0}{0.250} = 8.0\text{ M}, \quad [H_2S] = \frac{2.0}{0.250} = 8.0\text{ M}$$ $$[H_2] = \frac{1.0}{0.250} = 4.0\text{ M}, \quad [CS_2] = \frac{1.0}{0.250} = 4.0\text{ M}$$Substituting into the Qc expression:
$$Q_c = \frac{(4.0)^4(4.0)}{(8.0)(8.0)^2} = \frac{(256)(4.0)}{(8.0)(64)} = \frac{1024}{512} = 2.0$$c) Which side the equilibrium lies on: Since Qc = 2.0 is much greater than Kc = 0.036, the current mixture has far more products relative to reactants than the equilibrium would allow. The equilibrium therefore lies toward the reactants, the left side, at this temperature.
d) How the system will shift: Since Qc > Kc, the reaction must proceed in the reverse direction, converting some products back into reactants, in order to decrease Qc down to the value of Kc. The system will shift to the left, favoring CH₄ and H₂S, to reach equilibrium.
Question 14: Cysteine and Isoelectric Point
Cysteine has pKa₁ = 1.96, the carboxyl group, -COOH, and pKa₂ = 8.18, representing the group that governs the transition around neutral pH in this simplified two-pKa treatment.
a) i) Major product when cysteine reacts with NaOH(aq): Sodium hydroxide is a strong base, which deprotonates the acidic -COOH group of cysteine, forming the carboxylate salt, the sodium salt of cysteine, releasing water.
ii) Major product when cysteine reacts with HCl(aq): Hydrochloric acid is a strong acid, which protonates the basic -NH₂ group of cysteine, forming the corresponding ammonium chloride salt, the protonated amino group, -NH₃⁺, paired with Cl⁻.
b) Dominant structure of cysteine at the isoelectric point: At the isoelectric point, the amino acid exists predominantly as a zwitterion, meaning it carries both a positive charge, on the protonated amino group, -NH₃⁺, and a negative charge, on the deprotonated carboxyl group, -COO⁻, simultaneously, giving an overall net charge of zero.
c) Isoelectric point value for cysteine:
Step by step: For an amino acid whose net charge transitions from positive to neutral to negative across two relevant pKa values, the isoelectric point is calculated as the average of the two pKa values that bracket the neutral zwitterion form:
$$pI = \frac{pK_{a1} + pK_{a2}}{2} = \frac{1.96 + 8.18}{2} \approx 5.07$$Question 15: Electrophilic Substitution on Benzene Rings
a) Systematic IUPAC name of benzene: Benzene is officially named benzene itself under IUPAC nomenclature, since it is a retained trivial name for the parent aromatic ring; when used as a substituent group it is referred to as phenyl.
b) Necessary conditions for the reaction of methylbenzene with Br₂ to give the CH₂Br product: Since the product shows bromination occurring on the side-chain methyl group, forming -CH₂Br, rather than on the aromatic ring itself, this is a free-radical substitution, which requires ultraviolet (UV) light, or another radical initiator such as heat, rather than a halogen carrier catalyst like FeBr₃, which would instead direct bromination onto the ring.
c) Which substituent entered first, with reasoning:
i) The compound shows -NO₂ and -CH₃ groups in a para relationship (1,4-positions) on the ring. The methyl group (-CH₃) entered first. This is because -CH₃ is an activating, ortho/para-directing group, meaning if it were already on the ring, a second substitution would naturally occur at the para position, exactly as shown. If -NO₂, a deactivating, meta-directing group, had entered first, the second substituent would be directed to the meta position instead, not para, so the observed para relationship confirms methyl entered first.
ii) The compound shows -COOH and -CH₃ groups in a meta relationship (1,3-positions) on the ring. The carboxyl group (-COOH) entered first. -COOH is a deactivating, meta-directing group. If -COOH were already present on the ring, a second substituent would be directed to the meta position, exactly as shown. If -CH₃, an ortho/para director, had entered first, the second group would appear at the ortho or para position instead, not meta, confirming that -COOH was the first substituent.
Section B: Attempt Only Three Questions (30 marks)
Question 16: First-Order Reaction Kinetics
The reaction: 2A → B + C, first order, takes 30 minutes for 50% of the initial reactant to be consumed.
a) Define half-life: The half-life of a reaction is the time required for the concentration of a reactant to decrease to exactly half of its initial value. For a first-order reaction, the half-life is constant and independent of the starting concentration.
b) Half-life for this reaction: Since the question states that 50% of the reactant is consumed in 30 minutes, this is, by definition, the half-life. Answer: t₁/₂ = 30 minutes.
c) Rate constant k: For a first-order reaction, half-life and rate constant are related by:
$$t_{1/2} = \frac{\ln 2}{k} \implies k = \frac{\ln 2}{t_{1/2}} = \frac{0.693}{30} \approx 0.0231\ \text{min}^{-1}$$d) Time required for 90% of the initial reactant to react to completion: If 90% has reacted, 10% of the original concentration remains, so \([A]_t/[A]_0 = 0.10\). Using the integrated rate law:
$$\ln[A]_t = -kt + \ln[A]_0 \implies \ln\left(\frac{[A]_t}{[A]_0}\right) = -kt$$ $$\ln(0.10) = -(0.0231)t \implies t = \frac{-\ln(0.10)}{0.0231} = \frac{2.303}{0.0231} \approx 99.7\ \text{minutes}$$Answer: approximately 100 minutes.
e) True or false statements:
i) "A graph of [A] versus time is a straight line." This is False. For a first-order reaction, [A] decreases exponentially with time, producing a curve, not a straight line. It is a graph of ln[A] versus time that produces a straight line, with gradient −k, which is precisely why the integrated rate law is written in logarithmic form.
ii) "The rate of the reaction is minus one-half the rate of disappearance of A." This is True. Since the stoichiometry is 2A → B + C, two moles of A are consumed for every one mole of reaction that occurs. The overall reaction rate is therefore defined as rate = −(1/2)(d[A]/dt), meaning the reaction rate is indeed half the magnitude of the rate at which A disappears, matching the coefficient of A in the balanced equation.
Question 17: Bonding Trends in Period 3 Chlorides
The chlorides in question, with melting points: NaCl (1081 K), MgCl₂ (987 K), AlCl₃ (451 K, sublimes), SiCl₄ (203 K), PCl₅ (435 K), S₂Cl₂ (410 K).
a) Completing the bonding row:
Step by step: Moving across period 3, the elements become progressively less metallic and more non-metallic, which shifts the bonding character from purely ionic toward purely covalent. Sodium and magnesium are metals that form simple ionic lattices with very high melting points. Aluminium, at the boundary, forms a chloride with significant covalent character due to the high charge density of the small, highly charged Al³⁺ ion polarizing the chloride ion's electron cloud, which is why AlCl₃ sublimes rather than melting at a conventional high temperature. Silicon, phosphorus, and sulfur are non-metals that form simple molecular covalent chlorides, held together only by weak intermolecular, van der Waals, forces between molecules, giving them low melting points.
Answer:
NaCl: ionic | MgCl₂: ionic | AlCl₃: ionic with covalence | SiCl₄: covalent | PCl₅: covalent | S₂Cl₂: covalent
b) Equation for phosphorus reacting with chlorine to form PCl₅, with state symbols:
$$P_4(s) + 10Cl_2(g) \rightarrow 4PCl_5(s)$$c) Why the melting point of MgCl₂ is lower than that of NaCl: Although both are ionic compounds, Mg²⁺ has a higher charge and a smaller ionic radius than Na⁺, which would normally be expected to create a stronger ionic bond with greater lattice energy. However, this same high charge density on Mg²⁺ causes it to significantly polarize the electron cloud of the chloride ions, introducing a degree of covalent character into the bonding that weakens the purely ionic lattice structure compared to NaCl, which has almost pure ionic bonding. This partial covalent character is why MgCl₂ melts at a lower temperature than the more purely ionic NaCl, despite Mg²⁺ having a higher charge.
d) i) pH of a sodium chloride solution: Answer: b) Equal to 7. Justification: Na⁺ is the conjugate acid of a strong base, NaOH, and Cl⁻ is the conjugate base of a strong acid, HCl, so neither ion undergoes any significant hydrolysis reaction with water. NaCl is therefore a neutral salt, and its aqueous solution has a pH of exactly 7.
ii) The solution of aluminium chloride is: Answer: d) Acidic. Justification: As shown by the given equilibrium, the highly charged [Al(H₂O)₆]³⁺ ion is itself weakly acidic, since it can donate a proton to a surrounding water molecule, forming [Al(H₂O)₅OH]²⁺ and H₃O⁺, releasing hydronium ions into solution and lowering the pH below 7.
Question 18: Polymers F and G
Polymer F shows a repeating -CH₂-CHCl- backbone, and polymer G shows a repeating ester linkage with -(CH₂)₂- units connected by -C(=O)-O- groups.
a) Structural formulae of the monomers of F and G: The monomer of F is chloroethene (vinyl chloride), CH₂=CHCl. Polymer G is a polyester, and based on the repeating -(CH₂)₂-C(=O)-O- pattern, it is formed from the condensation of a diol and a diacid; the simplest matching monomer pair is ethylene glycol, HO-CH₂-CH₂-OH, and a suitable diacid such as ethanedioic (oxalic) acid, HOOC-COOH, though the exact diacid depends on precisely how many carbons separate the ester linkages in the original diagram.
b) Names of the constituent monomers: F is made from chloroethene (vinyl chloride) monomer units. G, being a polyester, is made from a diol and a dicarboxylic acid monomer, condensing together with loss of water at each linkage.
c) Repeating units: The repeating unit of F is written as [-CH₂-CHCl-]ₙ, reflecting the addition polymerization of the vinyl chloride double bond. The repeating unit of G is written as [-O-(CH₂)₂-O-CO-(CH₂)ₓ-CO-]ₙ, or the appropriate equivalent based on the exact diacid, reflecting the ester linkages formed by condensation.
d) Two applications of polymer F (polyvinyl chloride, PVC): PVC is widely used in plumbing pipes and fittings due to its durability and resistance to corrosion, and in electrical cable insulation, since it is a good electrical insulator and resistant to moisture.
e) Which of F or G was formed through condensation polymerization: Polymer G was formed through condensation polymerization. This is because G's backbone contains ester linkages, -C(=O)-O-, connecting the monomer units, which form when a diol and a diacid react together with the elimination of a small molecule, water, at each linkage point. Polymer F, by contrast, is formed by addition polymerization, where the C=C double bond in vinyl chloride simply opens up and monomers join directly to one another with no atoms lost, which is why F's backbone is a continuous carbon chain with no linking groups between repeat units.
Question 19: Born-Haber Cycle for Cesium Chloride
a) Define lattice energy: Lattice energy is the enthalpy change when one mole of an ionic solid is formed from its constituent gaseous ions, or, depending on convention, the energy required to separate one mole of an ionic solid completely into its gaseous ions.
b) Identifying enthalpy changes from the diagram:
i) Formation of cesium chloride: ΔH₁ = −433 kJ/mol, this is the overall enthalpy of formation, from Cs(s) + ½Cl₂(g) directly to CsCl(s).
ii) First ionization of cesium: ΔH₃ = +376 kJ/mol, the step converting gaseous Cs atoms into Cs⁺ ions plus an electron.
iii) First electron affinity of chlorine: ΔH₅ = −364 kJ/mol, the step where a gaseous chlorine atom gains an electron to form Cl⁻.
c) Calculating the lattice energy, ΔH₆:
Step by step: By Hess's Law, the direct path, ΔH₁, formation, must equal the sum of all the steps around the indirect cycle, ΔH₂ through ΔH₆:
$$\Delta H_1 = \Delta H_2 + \Delta H_3 + \Delta H_4 + \Delta H_5 + \Delta H_6$$Substituting the known values, ΔH₂ = +79, ΔH₃ = +376, ΔH₄ = +121, ΔH₅ = −364:
$$-433 = 79 + 376 + 121 + (-364) + \Delta H_6$$ $$-433 = 212 + \Delta H_6 \implies \Delta H_6 = -433 - 212 = -645\ \text{kJ/mol}$$Answer: ΔH₆ = −645 kJ/mol. The negative sign confirms this is the energy released when gaseous Cs⁺ and Cl⁻ ions come together to form the solid ionic lattice, consistent with lattice formation being an exothermic process.
d) Hess's diagram for C(s) + 2H₂(g) → CH₄(g): This diagram is constructed using an indirect route via complete combustion, since combustion enthalpies are experimentally measurable even when the direct formation reaction is not. The diagram places C(s) + 2H₂(g) and CH₄(g) at the top, connected directly by the unknown enthalpy of formation, ΔH°f CH₄(g). Both are then connected downward to a common set of combustion products, CO₂(g) + 2H₂O(l), at the bottom: C(s) + O₂(g) reaches this bottom level via ΔH°c C(s), 2H₂(g) + O₂(g) reaches it via 2×ΔH°c H₂(g), and CH₄(g) + 2O₂(g) reaches it via ΔH°c CH₄(g). Applying Hess's Law around this cycle allows ΔH°f CH₄(g) to be calculated as: ΔH°c C(s) + 2×ΔH°c H₂(g), minus ΔH°c CH₄(g).
Question 20: Acid-Base Indicators and Titration Curves
a) Using Le Châtelier's principle on the methyl orange equilibrium (H-MeO ⇌ MeO⁻ + H₃O⁺, red ⇌ yellow):
i) Adding methyl orange to HCl(aq): HCl is a strong acid, so it provides a high concentration of H₃O⁺ ions. By Le Châtelier's principle, this excess H₃O⁺, a product of the equilibrium, shifts the equilibrium to the left, favoring the formation of more H-MeO. The solution turns red, the color of the protonated, undissociated form.
ii) Adding methyl orange to NaOH(aq): NaOH is a strong base, so the OH⁻ ions react with and remove H₃O⁺ ions from the equilibrium. This shifts the equilibrium to the right, favoring the formation of more MeO⁻. The solution turns yellow, the color of the deprotonated form.
b) pH range of methyl orange, given pKᵢ = 3:
Step by step: An indicator typically shows a visible color transition across a pH range of approximately pKᵢ ± 1, since this is the range over which both the protonated and deprotonated forms are present in comparable, visually detectable proportions.
$$\text{pH range} = 3 \pm 1 = 2 \text{ to } 4$$c) Plotting the titration curve of 20 cm³ of 0.1M HCl titrated with 0.1M NaOH:
Plotting the given data, volume of NaOH added on the x-axis, pH on the y-axis, produces the classic strong acid to strong base titration curve shape: the pH rises slowly and gradually from pH 1.00 up to around pH 2.30 as the first 18 mL of NaOH is added, since a large excess of unreacted HCl is still present to resist the pH change. Between 18 mL and 21 mL, the curve rises extremely steeply, climbing from around pH 2.30 all the way up to pH 11.38 in just a few milliliters, because this narrow volume range straddles the equivalence point at 20 mL, where moles of NaOH added exactly equal the original moles of HCl, giving pH = 7.00 exactly at that point. Beyond 21 mL, the curve flattens out again into a gentle upward slope as excess NaOH accumulates, reaching pH 12.43 by 35.5 mL. This steep vertical jump centered on the 20 mL equivalence point is the key feature to include when sketching the curve, and it is also why an indicator like methyl orange, with its pH 2 to 4 transition range, or phenolphthalein, with a higher transition range, can each be used successfully for this particular strong acid to strong base titration, since the pH jump at the equivalence point spans a very wide range that comfortably covers either indicator's transition window.
Study Tips for S6 Chemistry II
This paper rewards recognizing recurring reasoning patterns rather than memorizing isolated facts: predicting bonding character from periodic trends in Question 17, predicting acidity or basicity of a salt solution from the strength of its parent acid and base in Questions 17 and 20, and using Le Châtelier's principle to predict shifts in any equilibrium, whether it's an indicator color change or a gas-phase reaction, in Questions 13 and 20, are all the same underlying skill applied to different chemical contexts. Practicing that transferable reasoning, rather than memorizing this paper's specific answers, will serve you well across chemistry topics you haven't seen before.
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