CSEC Chemistry — Paper 2 Solutions

Read the level of mercury in each thermometer:

1 mark per correct reading from each of the five thermometers.

On Figure 2 (page 7): time (s) on the x-axis from 0 to 120; temperature (°C) on the y-axis from 0 to 50.

Plot the 9 points: (0, 25), (15, 33), (30, 38), (45, 43), (60, 44.5), (75, 46), (90, 45.5), (105, 45.5), (120, 45).

Draw a single smooth curve. The temperature rises sharply between 0 and ~75s as the reaction releases heat, reaches a peak (~46°C) around 75 s, then falls slowly as the calorimeter loses heat to the surroundings.

Marks: 1 for axes labelled with units; 1 for sensible scale that uses most of the grid; 2 for plotting all points correctly; 1 for a smooth best-fit curve.

The "true" temperature change ignores the heat lost to the surroundings during the reaction. Extrapolate the cooling part of the curve (after the peak) back to t = 0 as a straight line and read the temperature there — that gives you the maximum temperature the mixture would have reached if no heat was lost. Then ΔT = (extrapolated maximum) − (initial temperature).

Reading from the graph: extrapolated maximum ≈ 47°C, initial = 25°C.

ΔT ≈ 47 − 25 = 22°C (accept 21–23 °C).

An exothermic reaction releases heat to the surroundings (temperature of the surroundings rises; ΔH is negative). An endothermic reaction absorbs heat from the surroundings (temperature of the surroundings falls; ΔH is positive).

1 mark for direction of heat flow; 1 mark for sign of ΔH or effect on surroundings.

The temperature of the mixture rose from 25°C to ~47°C — the reaction released heat to the solution.

Exothermic.

Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g)

1 mark for correct reactants/products; 1 mark for balancing (the 2 in front of HCl); 1 mark for all four state symbols correct.

Moles = mass ÷ RAM = 0.48 g ÷ 24 g/mol

= 0.020 mol of Mg (i.e. 2.0 × 10⁻² mol).

From the equation: 1 mol Mg → 1 mol H₂. So moles of H₂ produced = moles of Mg = 0.020 mol. (1 mark)

Volume at STP = moles × molar volume = 0.020 × 22 400 cm³ (1 mark)

Volume of H₂ at STP = 448 cm³ (1 mark)

Volume of solution = 100 cm³, density = 1 g/cm³ ⇒ mass m = 100 g. (1 mark)

ΔH = m × c × ΔT = 100 × 4.2 × 22 (1 mark)

ΔH = 9 240 J = 9.24 kJ (energy released) (1 mark)

Sign: this is heat released, so ΔH = −9.24 kJ.

ΔH per mole of Mg = total energy ÷ moles of Mg = 9.24 kJ ÷ 0.020 mol

ΔH = −462 kJ/mol of Mg (negative because the reaction is exothermic).

An alloy is a mixture of a metal with one or more other elements (usually other metals or carbon), formed to give properties superior to those of the pure metal.

• Stainless steel (iron + chromium + nickel)
• Mild steel / carbon steel (iron + small percentage of carbon)
• Other accepted: tool steel, high-speed steel, manganese steel.

1 mark per correct type.

1. Greater hardness / strength — steel is harder and stronger than pure iron.
2. Better resistance to corrosion / rusting (especially stainless steel).
3. Other accepted: more durable; tougher / less brittle; can be tempered to suit different uses.

1 mark each.

The pattern of thermal decomposition is the key to reactivity:

• Less-reactive metal nitrates decompose to give the metal oxide + nitrogen dioxide + oxygen (e.g. Cu(NO₃)₂, Pb(NO₃)₂).
• Very reactive metal nitrates (Group I metals, e.g. Na, K) decompose only as far as the metal nitrite + oxygen.

More reactive metals: W and Z (their nitrates → metal nitrite + O₂). Less reactive metals: V, X and Y (their nitrates → metal oxide + NO₂ + O₂).

1 mark for correctly identifying the more-reactive group; 1 mark for the less-reactive group.

Sodium nitrate, like potassium nitrate, decomposes to give sodium nitrite and oxygen only:

2NaNO₃(s) → 2NaNO₂(s) + O₂(g)

1 mark for correct products; 1 mark for balancing and state symbols.

Use displacement: a more-reactive metal will displace a less-reactive metal from its oxide.

• V did not react with any of the other oxides ⇒ V is the least reactive.
• X reacted with the oxide of V and with the oxide of Y ⇒ X is more reactive than both V and Y.
• Y reacted only with the oxide of V ⇒ Y is more reactive than V but less reactive than X.

Order, most → least reactive: X > Y > V.

1 mark for correctly placing X first; 1 mark for the full correct order.

2 marks for correctly identifying Fe²⁺ + NH₄⁺; 2 marks for the supporting reasoning. Bonus equation: NH₄⁺(aq) + OH⁻(aq) → NH₃(g) + H₂O(l).

Compound A: CH₃−CH₂−COOH (it has a −COOH group at the end).

Carboxylic acids (alkanoic acids).

Carboxyl group, −COOH (i.e. −C(=O)−O−H).

Propanoic acid (CH₃CH₂COOH, three carbons including the −COOH carbon).

A carboxylic acid + an alcohol → ester + water (a small molecule released).

Condensation polymerisation (water is the small molecule eliminated each time).

Compound B: H₂C=CH−CH₂−CH₃ (contains a C=C double bond).

Alkenes.

Carbon–carbon double bond, C=C (the alkene functional group).

Four carbons, double bond between C1 and C2.

But-1-ene (CH₂=CHCH₂CH₃).

Structural isomers are compounds that have the same molecular formula but different structural formulae — i.e. the same numbers of each atom, but the atoms are arranged differently (different connectivity, branching or position of a functional group).

1 mark for "same molecular formula"; 1 mark for "different structural formula / arrangement".

The other two pentane isomers are 2-methylbutane and 2,2-dimethylpropane:

Isomer 1 — 2-methylbutane (2 marks)

       CH3
       |
H3C − CH − CH2 − CH3       (fully displayed: every C, H drawn)

   H   H   H   H
   |   |   |   |
H− C − C − C − C −H
   |   |   |   |
   H  CH3  H   H

Isomer 2 — 2,2-dimethylpropane (neopentane) (2 marks)

           CH3
           |
   H3C − C − CH3
           |
           CH3

The central C is bonded to four methyl groups; each
CH3 has 3 H atoms and the central C has none of its own
hydrogens.
      

2 marks each for fully displayed structure (must show every C, H and bond explicitly).

• Isomer 1 — 2-methylbutane (1 mark)
• Isomer 2 — 2,2-dimethylpropane (also called neopentane) (1 mark)

The copper electrodes are ACTIVE. (1 mark)

Reason: The copper electrodes take part in the electrode reactions — at the anode, copper is oxidised (Cu dissolves into solution as Cu²⁺), and at the cathode Cu²⁺ is reduced and copper deposits on it. Inert electrodes (e.g. graphite, platinum) would not dissolve. (1 mark)

• Cathode (− electrode): positively charged cations migrate to it; they gain electrons from the electrode (reduction) and form atoms or molecules. (1 mark)
• Anode (+ electrode): negatively charged anions migrate to it; they lose electrons to the electrode (oxidation) and form atoms or molecules. (Active electrodes can themselves lose electrons.) (1 mark)

The battery's + terminal is wired to electrode A and the − terminal is wired to electrode B.

• Electrode A — anode (connected to the + terminal). (1 mark)
• Electrode B — cathode (connected to the − terminal). (1 mark)

CuSO₄(aq) contains Cu²⁺, SO₄²⁻ ions, plus H⁺ and OH⁻ from water.

• Ions at cathode (positive ions): Cu²⁺ and H⁺. (1 mark)
• Ions at anode (negative ions): SO₄²⁻ and OH⁻. (1 mark)

Cu²⁺ is preferentially discharged at the cathode (lower in the electrochemical series than H⁺). With copper anodes, the copper itself is oxidised in preference to OH⁻/SO₄²⁻.

• Cathode (reduction): Cu²⁺(aq) + 2e⁻ → Cu(s) (1 mark)
• Anode (oxidation): Cu(s) → Cu²⁺(aq) + 2e⁻ (1 mark)

Redox = simultaneous reduction + oxidation.

At the cathode Cu²⁺ ions gain electrons → reduced to Cu metal. At the anode Cu atoms lose electrons → oxidised to Cu²⁺. Reduction at one electrode + oxidation at the other = a redox reaction.

1 mark for stating reduction at cathode; 1 mark for stating oxidation at anode.

The cathode increases in mass as a fresh layer of copper metal is deposited on it (it becomes thicker / coated with shiny brown-pink copper).

1. Replace the copper cathode with the steel object — connect the piece of steel to the negative terminal of the battery so it acts as the cathode. (1 mark)
2. Keep a pure copper bar as the anode in CuSO₄(aq) solution. (1 mark)

When current flows: Cu dissolves from the copper anode → Cu²⁺ in solution → Cu metal plates onto the steel cathode. Use a low, steady current and clean the steel first to ensure an even, well-bonded plate.

1. Same general formula for every member of the series.
2. Same functional group — so members show similar chemical properties.
3. Successive members differ by one −CH₂− unit (i.e. by 14 g/mol of relative molecular mass).
4. Gradation in physical properties as molar mass increases (e.g. melting/boiling points rise; volatility falls; viscosity rises).

1 mark each = 4 marks.

General formula of alcohols: C_nH_2n+1OH (or C_nH_2n+2O). (1 mark)

5th member: n = 5 ⇒ C₅H₁₁OH.

Molecular formula of the 5th member: C₅H₁₂O (pentanol) (1 mark)

Compound D is shown end-to-end as: HO−C(=O)−CH₂−CH₂−CH=CH−CH₂−C(=O)−O−CH₂−CH₃.

Three functional groups are visible — circle any TWO:

1. Carboxyl group (−COOH) at the left end (−C(=O)−O−H).
2. Carbon–carbon double bond (C=C) in the middle (alkene).
3. Ester linkage (−COO−) further along (−C(=O)−O−CH₂−).

1 mark per correctly circled group.

Alkanes are saturated (only C−C single bonds), whereas alkenes are unsaturated and contain at least one C=C double bond.

E decolorises bromine water (orange → colourless) ⇒ E is an alkene.

F gives no reaction with bromine water (under normal conditions) ⇒ F is an alkane.

• Hydrocarbon E = ethene (C₂H₄) (1 mark)
• Hydrocarbon F = ethane (C₂H₆) (1 mark)

In the presence of UV light (sunlight) — alkanes undergo a free-radical substitution reaction with Br₂ only when irradiated by UV light, producing bromoalkane + HBr.

Ethene burns in excess oxygen to form carbon dioxide and water:

C₂H₄(g) + 3O₂(g) → 2CO₂(g) + 2H₂O(l)

1 mark for correct reactants/products; 1 mark for balancing (3 in front of O₂; 2's in front of CO₂ and H₂O); 1 mark for state symbols.

1. Sulfur dioxide (SO₂) — from burning sulfur-containing fuels; causes acid rain.
2. Nitrogen oxides (NO_x) — from high-temperature combustion in vehicles and factories; cause smog and acid rain.
3. Other accepted: CO (carbon monoxide), CO₂ (greenhouse gas), volatile organic compounds (VOCs), particulate matter, CFCs.

1 mark per pollutant.

The factory makes pesticides, fertilisers and plastics. Algal blooms (eutrophication) are caused by excess nutrients in the water — chiefly:

1. Nitrate ions (NO₃⁻) from nitrogen fertilisers.
2. Phosphate ions (PO₄³⁻) from phosphate fertilisers.
3. Ammonium ions (NH₄⁺) from ammonium-based fertilisers (or from spilled ammonia).

1 mark per substance.

Green chemistry is the design of chemical products and processes that reduce or eliminate the use and generation of hazardous substances — i.e. chemistry that is safer for human health and the environment.

1. Prevention of waste — design reactions so that little or no waste is produced (better than treating waste afterwards).
2. Atom economy — design reactions so that as many of the reactant atoms as possible end up in the final product.
3. Use of renewable feedstocks — use raw materials from renewable biological sources rather than non-renewable petroleum where possible.
4. Other accepted: use safer solvents/auxiliaries; design for energy efficiency; design safer chemicals/products; design for biodegradability; use catalysts rather than stoichiometric reagents; minimise derivatives.

1 mark per correctly stated principle.

1. Prevention of waste: Redesign reactions so by-products are minimised; recycle solvents and catalysts internally rather than discharging them. Less waste means less SO₂ and less fertiliser run-off into the surrounding farmland and rivers. (2 marks)
2. Use of safer chemicals / safer solvents: Replace toxic pesticide intermediates with safer biodegradable alternatives, and replace volatile organic solvents with water or supercritical CO₂. This reduces the toxic burden on plant and animal life nearby. (2 marks)
3. Design for biodegradability: Make plastics that are biodegradable (e.g. PLA from corn starch) so any waste plastic that escapes the factory does not persist in the environment as non-biodegradable solid waste. (2 marks)
4. Other accepted points: use catalysts instead of stoichiometric reagents; capture and reuse exhaust gases; install scrubbers to remove SO₂ and NO_x; use renewable plant-based feedstocks; design products that decompose to harmless substances after use.

2 marks for each principle properly applied to the factory's situation = 6 marks.