Level 4 — ApplicationGreen Chemistry & Sustainability

Green Chemistry & Sustainability

⏱ 60 minutes60 marksprintable — key stays hidden on paper

Level 4 — Application (novel/unseen problems, no hints) Time limit: 60 minutes Total marks: 60

Answer all questions. Show all working. Give atomic masses to 1 decimal place where needed (H = 1.0, C = 12.0, N = 14.0, O = 16.0, Na = 23.0, Cl = 35.5, Br = 79.9, Fe = 55.8).

Question 1 — Atom economy of a new synthesis route (14 marks)

A chemist proposes making the drug intermediate allyl chloride (C₃H₅Cl, M = 76.5) by two competing routes.

Route A (substitution): $C_3H_5OH + HCl \rightarrow C_3H_5Cl + H_2O$

Route B (addition–elimination waste route): $C_3H_6 + Cl_2 \rightarrow C_3H_5Cl + HCl$

(a) Calculate the percentage atom economy of each route with respect to allyl chloride. (6)

(b) Route A gives 82% yield; Route B gives 95% yield. Calculate the effective yield of desired atoms per unit reactant mass by combining atom economy with percentage yield for each route, and state which route is greener on this combined basis. (5)

(c) Route B's byproduct HCl can be captured and resold. Explain, referencing a specific principle of green chemistry, how this changes your assessment and what waste metric would better capture it than atom economy alone. (3)

Question 2 — Solvent selection for an unseen reaction (12 marks)

A company must extract a non-polar terpene from plant biomass and later run a hydrolysis (needs water) on the same product line.

(a) The company shortlists three solvents: liquid water, supercritical CO₂ (scCO₂), and an imidazolium ionic liquid. For the extraction step, rank the three by expected effectiveness for a non-polar solute and justify each ranking in one sentence. (6)

(b) scCO₂ is chosen for extraction. Given CO₂'s critical point is $T_c = 31\,°C$ , $P_c = 73.8\ \text{bar}$ , explain two practical process advantages over ionic liquids for a heat-sensitive terpene. (4)

Question 3 — Green propellant performance (12 marks)

A CubeSat mission compares hydrazine (N₂H₄) with the green monopropellant AF-M315E (hydroxylammonium nitrate based).

(a) Hydrazine decomposes catalytically: $3N_2H_4 \rightarrow 4NH_3 + N_2$ Calculate the mass of N₂ produced per 100 g of hydrazine decomposed. (4)

(b) AF-M315E offers ~12% higher specific impulse ( $I_{sp}$ ) and ~65% higher density than hydrazine. Explain quantitatively why the density × I_sp ("density impulse") metric is more relevant than I_sp alone for a small-volume CubeSat, and estimate the combined density-impulse advantage factor. (5)

(c) Give two operational (non-performance) green advantages of AF-M315E over hydrazine that reduce launch-processing cost. (3)

Question 4 — Hydrogen economy round-trip efficiency (12 marks)

A solar-to-hydrogen-to-electricity storage system uses electrolysis then a fuel cell.

(a) Water electrolysis: $2H_2O \rightarrow 2H_2 + O_2$ . Calculate the mass of H₂ produced from electrolysing 1.00 kg of water at 100% Faradaic efficiency. (3)

(b) The electrolyser is 70% energy-efficient, gas compression/storage loses a further 10%, and the fuel cell is 55% efficient. Calculate the overall round-trip energy efficiency (electricity-in to electricity-out). (4)

(c) A competing scheme stores the same solar electricity in a lithium battery at 90% round-trip efficiency. Give two reasons a mission designer might still choose the hydrogen route despite its lower efficiency. (3)

(d) State one green-chemistry justification for capturing the O₂ byproduct of electrolysis rather than venting it. (2)

Question 5 — Carbon capture mass balance (10 marks)

An amine scrubber captures CO₂ using monoethanolamine (MEA, C₂H₇NO, M = 61.0) in a 2:1 MEA:CO₂ carbamate reaction: $2\,C_2H_7NO + CO_2 \rightarrow \text{carbamate salt}$

(a) Calculate the mass of MEA required to capture 1.00 tonne of CO₂. (4)

(b) Regenerating the amine costs 3.7 GJ per tonne CO₂ captured. If the captured CO₂ came from burning methane ( $CH_4 + 2O_2 \rightarrow CO_2 + 2H_2O$ , ΔH = −890 kJ/mol) to generate the regeneration energy, calculate what fraction of the methane's combustion energy is consumed just regenerating the amine per tonne of CO₂ captured. Comment on the sustainability implication. (6)

Answer keyMark scheme & solutions

Question 1 (14 marks)

(a) Atom economy = (M of desired product / Σ M of all products) × 100.

Route A: products C₃H₅Cl (76.5) + H₂O (18.0). Total = 94.5. AE = 76.5/94.5 × 100 = 80.95% ≈ 81.0% (3 marks)

Route B: products C₃H₅Cl (76.5) + HCl (36.5). Total = 113.0. AE = 76.5/113.0 × 100 = 67.7% (3 marks)

(b) Combined metric = AE × yield:

Route A: 0.8095 × 0.82 = 0.664 (66.4%) (2)
Route B: 0.677 × 0.95 = 0.643 (64.3%) (2)

Route A is marginally greener on this combined basis (66.4% vs 64.3%). (1)

(c) (3 marks) By capturing and reselling HCl, the byproduct is no longer waste — this invokes Principle 1 (waste prevention) and Principle 2 (atom economy)/waste valorisation. Atom economy penalises HCl as "lost" atoms, but if it is a co-product with value, the true environmental picture is better captured by the E-factor (mass of actual waste / mass of product), which counts only genuinely discarded material. Award 1 for principle, 1 for E-factor, 1 for reasoning.

Question 2 (12 marks)

(a) (6 marks — 2 each)

scCO₂ — best: non-polar/low-polarity supercritical fluid dissolves non-polar terpenes well; tunable density.
Ionic liquid — intermediate: can be tuned, but imidazolium ILs are fairly polar/ionic so dissolve non-polar terpenes only moderately.
Water — worst: highly polar, will not dissolve a non-polar terpene appreciably.

(b) (4 marks — 2 each) (i) Mild critical temperature (31 °C) means extraction occurs near ambient, protecting the heat-sensitive terpene, whereas ILs may require heating to lower viscosity. (ii) scCO₂ is removed simply by depressurisation, leaving no solvent residue in the product; ILs are non-volatile and hard to separate cleanly from product.

(c) (2 marks) ILs are often toxic, poorly biodegradable, and energy-intensive/expensive to synthesise — so despite "non-volatile ⇒ low air emissions," their aquatic toxicity or synthetic footprint can make them non-green.

Question 3 (12 marks)

(a) M(N₂H₄) = 32.0. 100 g → 100/32.0 = 3.125 mol N₂H₄. Ratio: 3 N₂H₄ → 1 N₂, so mol N₂ = 3.125/3 = 1.0417 mol. Mass N₂ = 1.0417 × 28.0 = 29.2 g (4 marks: mol hydrazine 1, ratio 1, mol N₂ 1, mass 1)

(b) (5 marks) A CubeSat is volume-limited, not mass-limited, for propellant tankage. Density impulse = ρ × I_sp measures thrust-time capability per unit tank volume. Combined factor = 1.12 (I_sp) × 1.65 (density) = 1.85, i.e. ~85% more usable impulse per unit volume than hydrazine. (2 for reasoning, 2 for factor calc, 1 for value)

(c) (3 marks — any 2, 1.5 each ≈ 3) Lower toxicity/vapour pressure ⇒ no SCAPE suits or exclusion zones during fuelling; not acutely toxic/carcinogenic so simpler handling permits; higher storage stability/less stringent transport classification.

Question 4 (12 marks)

(a) M(H₂O) = 18.0. 1000 g / 18.0 = 55.56 mol H₂O → 55.56 mol H₂ (2 H₂O → 2 H₂, 1:1). Mass H₂ = 55.56 × 2.0 = 111 g (3 marks)

(b) Overall η = 0.70 × 0.90 × 0.55 = 0.3465 ≈ 34.7% (4 marks: identify multiplicative chain 2, compute 2)

(c) (3 marks — any 2) Hydrogen offers far higher gravimetric energy density (mass-critical long missions), no self-discharge over long storage, decoupled energy/power scaling (large seasonal storage cheaper), and H₂ can also serve as a chemical feedstock/propellant.

(d) (2 marks) O₂ is a valuable co-product; capturing it avoids waste (Principle 1) and displaces energy-intensive cryogenic air separation elsewhere, improving overall process atom/energy economy.

Question 5 (10 marks)

(a) 1.00 t CO₂ = 1.00×10⁶ g. M(CO₂) = 44.0 → 22,727 mol CO₂. MEA needed = 2 × 22,727 = 45,455 mol. Mass = 45,455 × 61.0 = 2.77×10⁶ g = 2.77 tonnes MEA (4 marks: mol CO₂ 1, 2:1 ratio 1, mol MEA 1, mass 1)

(b) Per tonne CO₂: 22,727 mol CO₂ produced. From CH₄ combustion, 1 mol CH₄ → 1 mol CO₂ releasing 890 kJ. Energy released producing that CO₂ = 22,727 × 890 kJ = 2.023×10⁷ kJ = 20.2 GJ. Regeneration cost = 3.7 GJ. Fraction = 3.7 / 20.2 = 0.183 ≈ 18% (4 marks: mol 1, energy 1.5, fraction 1.5)

Comment (2 marks): ~18% of the fuel's energy is consumed just running capture (an energy penalty/parasitic load); this lowers net efficiency and shows carbon capture is only sustainable if the regeneration energy itself is low-carbon (renewable), else you burn more fuel — and emit more CO₂ — to capture it.

[
{"claim":"Route A atom economy 80.95%","code":"ae=76.5/94.5*100; result = abs(ae-80.95)<0.05"},
{"claim":"Route B atom economy 67.7%","code":"ae=76.5/113.0*100; result = abs(ae-67.70)<0.05"},
{"claim":"N2 mass from 100g hydrazine = 29.2 g","code":"m=(100/32.0)/3*28.0; result = abs(m-29.17)<0.1"},
{"claim":"H2 from 1kg water = 111 g","code":"m=(1000/18.0)*2.0; result = abs(m-111.1)<0.5"},
{"claim":"MEA mass to capture 1 t CO2 = 2.77 t","code":"m=2*(1e6/44.0)*61.0; result = abs(m-2.77e6)<1e4"},
{"claim":"Regeneration fraction of methane energy ~18%","code":"frac=3.7/((1e6/44.0)*890/1e6); result = abs(frac-0.183)<0.005"}
]