Level 3 — ProductionGreen Chemistry & Sustainability

Green Chemistry & Sustainability

⏱ 45 minutes60 marksprintable — key stays hidden on paper

Level 3 — Production (from-scratch derivations, explain-out-loud reasoning) Time limit: 45 minutes Total marks: 60

Instructions: Answer all questions. Show full working for all calculations. Use $...$ for inline math. Relative atomic masses: H = 1.008, C = 12.011, N = 14.007, O = 15.999, Na = 22.990, Cl = 35.45.

Question 1 — Atom economy from scratch (12 marks)

Consider the industrial synthesis of ethylene oxide by two routes:

Route A (chlorohydrin, older): $C_2H_4 + Cl_2 + H_2O \rightarrow C_2H_4(OH)Cl + HCl$ $C_2H_4(OH)Cl + Ca(OH)_2 \rightarrow 2\,C_2H_4O + CaCl_2 + 2H_2O \quad(\text{balance per mole of product below})$

Route B (direct oxidation): $C_2H_4 + \tfrac{1}{2}O_2 \rightarrow C_2H_4O$

(a) Define % atom economy and write its formula. (2) (b) For Route B, calculate the % atom economy for ethylene oxide ( $C_2H_4O$ ). (3) (c) For Route A, treating the overall two-step process (combine both equations so one mole of $C_2H_4$ gives one mole of $C_2H_4O$ ), calculate the overall % atom economy. Show the combined balanced equation. (5) (d) State which green chemistry principle atom economy most directly embodies, and explain in one sentence why Route B is greener. (2)

Question 2 — The 12 principles, applied (10 marks)

A pharmaceutical process is described:

"A batch reaction runs at 180 °C using a chlorinated solvent (dichloromethane) as reaction medium. It uses a stoichiometric heavy-metal oxidant, gives 40% yield, and the excess solvent is distilled and incinerated. Product purity is checked by taking hourly samples for HPLC."

Identify five distinct violations of the 12 principles of green chemistry in this process. For each, (i) name the principle violated and (ii) propose a concrete greener alternative. (10, i.e. 2 per point)

Question 3 — Solvent selection reasoning (10 marks)

(a) Supercritical CO₂ (scCO₂) is used as a green solvent. State its approximate critical temperature and pressure, and explain from scratch two physical reasons why the supercritical state makes it an attractive solvent/extraction medium. (5) (b) Ionic liquids are marketed as "green solvents," but this label is contested. Give two properties that support the green claim and two reasons the claim is questioned. (3) (c) Explain why water, despite being the greenest solvent by cost and toxicity, is often unsuitable for organic reactions. Give two reasons. (2)

Question 4 — Green propellants (10 marks)

Hydrazine ( $N_2H_4$ ) has long been the standard monopropellant.

(a) Explain from a molecular/toxicological standpoint two reasons hydrazine is being replaced. (3) (b) LMP-103S is based on ammonium dinitramide (ADN, $NH_4N(NO_2)_2$ ) and AF-M315E on hydroxylammonium nitrate (HAN, $NH_3OH^+NO_3^-$ ). State two performance advantages these ionic-liquid propellants have over hydrazine beyond toxicity. (4) (c) Give one practical operational challenge engineers face when switching from hydrazine to these green propellants. (3)

Question 5 — Hydrogen economy & electrolysis (12 marks)

(a) Write the balanced half-equations and the overall equation for the electrolysis of water in acidic conditions. Label anode and cathode. (4) (b) The standard cell potential for water splitting is $E^\circ = -1.23\ \text{V}$ . Using $\Delta G^\circ = -nFE^\circ$ with $F = 96485\ \text{C mol}^{-1}$ , calculate the minimum energy (in kJ) required to electrolyse one mole of water. State $n$ . (4) (c) A hydrogen fuel cell reverses this reaction. Write its overall reaction and state why the fuel cell is described as producing "zero emissions at point of use." (2) (d) Explain the distinction between "grey," "blue," and "green" hydrogen. (2)

Question 6 — Carbon capture, integrative (6 marks)

(a) Amine scrubbing captures $CO_2$ via reversible reaction with e.g. monoethanolamine. Explain why reversibility is the essential design feature, and where the captured $CO_2$ can go (name two fates). (3) (b) Carbon capture is energetically costly. Explain briefly why this creates a thermodynamic/sustainability tension, referencing the energy penalty. (3)

Answer keyMark scheme & solutions

Question 1 (12 marks)

(a) Atom economy measures the proportion of reactant mass that ends up in the desired product. (1) $\%\text{ AE} = \frac{\text{molar mass of desired product}}{\text{sum of molar masses of all reactants}} \times 100 \quad\textbf{(1)}$

(b) Route B: Reactants: $C_2H_4$ ( $M = 28.054$ ) $+ \tfrac12 O_2$ ( $M = 15.999$ ).

$M(C_2H_4) = 2(12.011)+4(1.008) = 24.022 + 4.032 = 28.054$ (1)
Total reactant mass $= 28.054 + 15.999 = 44.053$ ; $M(C_2H_4O)=28.054+15.999=44.053$ (1) $\%\text{AE} = \frac{44.053}{44.053}\times100 = 100\% \quad\textbf{(1)}$ (Direct oxidation adds only the O that appears in product → 100% AE.)

(c) Route A combined equation. Adding the two steps and cancelling the chlorohydrin intermediate. For one mole of $C_2H_4O$ : $C_2H_4 + Cl_2 + H_2O + \tfrac12 Ca(OH)_2 \rightarrow C_2H_4O + HCl + \tfrac12 CaCl_2 + H_2O$ Net (cancel one $H_2O$ ): $C_2H_4 + Cl_2 + \tfrac12 Ca(OH)_2 \rightarrow C_2H_4O + HCl + \tfrac12 CaCl_2 \quad\textbf{(2)}$ Reactant masses:

$C_2H_4 = 28.054$
$Cl_2 = 2(35.45) = 70.90$
$\tfrac12 Ca(OH)_2 = \tfrac12(40.078 + 2(15.999)+2(1.008)) = \tfrac12(74.092) = 37.046$ (1)

Total reactants $= 28.054 + 70.90 + 37.046 = 136.00$ $\%\text{AE} = \frac{44.053}{136.00}\times100 = 32.4\% \quad\textbf{(2)}$

(d) Embodies Principle 2 (Atom Economy) — maximise incorporation of all materials into the product. (1) Route B is greener because nearly all reactant atoms end up in product (100% vs 32%), producing no salt/HCl by-products/waste. (1)

Question 2 (10 marks)

Award 1 mark for correctly naming principle, 1 mark for a valid greener fix. Any five of:

Violation	Principle	Greener alternative
High temperature 180 °C	P6 Design for Energy Efficiency	Use catalyst / lower-temp route to reduce energy
Dichloromethane (chlorinated, toxic, volatile)	P5 Safer Solvents & Auxiliaries	Use water / scCO₂ / bio-based solvent
Stoichiometric heavy-metal oxidant	P9 Catalysis (not stoichiometric reagents) & P3 Less Hazardous Synthesis	Catalytic oxidation with $O_2/H_2O_2$
40% yield (much waste)	P2 Atom Economy / P1 Prevention of waste	Optimise route to higher yield
Solvent incinerated (not recovered)	P1 Prevention / P5	Recover & recycle solvent
Hourly sampling for HPLC (offline)	P11 Real-time analysis for pollution prevention	In-line/real-time process monitoring
Heavy metal = persistent	P3 Less hazardous / P10 Degradation	Non-toxic reagents

(2 marks × 5 = 10)

Question 3 (10 marks)

(a) scCO₂ critical point: $T_c \approx 31\ °C$ (304 K), $P_c \approx 74\ \text{bar}$ (~73 atm). (1) Reasons: (2 each, any two)

Gas-like diffusivity/low viscosity + liquid-like density → penetrates matrices fast and dissolves solutes efficiently (superior mass transfer for extraction). (2)
Solvent power tunable with pressure/temperature — small P/T changes vary density and hence dissolving power, and CO₂ is easily removed by depressurisation leaving no solvent residue (non-toxic, non-flammable, recyclable). (2)

(b) Green support (any two, ½ each): negligible vapour pressure → no VOC emission; non-flammable; thermally stable; recyclable/tunable. (1) Contested (any two, 1 each): many ILs are toxic/poorly biodegradable; energy-intensive & expensive to synthesise and purify; some contain fluorinated anions producing hazardous by-products. (2)

(c) Any two: (1 each) many organic reactants are insoluble/hydrophobic in water; water can hydrolyse/decompose reagents or intermediates (e.g. organometallics, acid chlorides); water's high boiling point/heat of vaporisation makes product isolation and drying energy-costly.

Question 4 (10 marks)

(a) Any two, 1.5 each (≈3): hydrazine is highly toxic/carcinogenic and readily absorbed (skin/inhalation); it is volatile and corrosive, requiring hazardous SCAPE suit handling and a large exclusion zone, raising ground-processing cost/risk. (3)

(b) Any two performance advantages (2 each):

Higher density and higher specific impulse (Isp) / density-Isp → more thrust/propellant volume, smaller tanks. (2)
Lower freezing point / higher storage stability → less heater power needed on spacecraft; also higher energy density. (2)

(c) Any one (3 marks): higher combustion/decomposition temperatures (~1600–1900 °C for HAN/ADN vs ~600–900 °C for hydrazine) demand new high-temperature catalyst bed and chamber materials; catalyst-bed preheating requirements; qualification/heritage risk for new hardware.

Question 5 (12 marks)

(a) Acidic electrolysis: (1 each label + eqns = 4)

Cathode (reduction): $2H^+ + 2e^- \rightarrow H_2$
Anode (oxidation): $2H_2O \rightarrow O_2 + 4H^+ + 4e^-$
Overall: $2H_2O \rightarrow 2H_2 + O_2$

(b) Per mole of water, overall $2H_2O \rightarrow 2H_2 + O_2$ transfers 4 electrons for 2 water → $n = 2$ per mole of water. (1) $\Delta G^\circ = -nFE^\circ = -(2)(96485)(-1.23) = +237{,}353\ \text{J} \approx +237\ \text{kJ} \quad\textbf{(3)}$ Minimum energy $\approx 237\ \text{kJ mol}^{-1}$ (endergonic → energy input required).

(c) Fuel cell overall: $2H_2 + O_2 \rightarrow 2H_2O$ (or $H_2 + \tfrac12O_2 \rightarrow H_2O$ ). (1) Only product is water, so no $CO_2$ /pollutants emitted where the cell operates → "zero emissions at point of use." (1)

(d) (2, any correct) Grey H₂ = from fossil fuels (steam methane reforming) releasing $CO_2$ ; Blue H₂ = same but with carbon capture & storage; Green H₂ = from electrolysis powered by renewable electricity (no net $CO_2$ ). (2)

Question 6 (6 marks)

(a) Reversibility is essential because the amine must bind $CO_2$ at the absorber (low T) then release it on heating in the stripper, regenerating the amine for reuse; without reversibility the sorbent would be single-use, making the process uneconomic and wasteful. (2) Fates of captured $CO_2$ (any two, ½ each = 1): geological storage/sequestration; enhanced oil recovery; conversion to fuels/chemicals; carbonate feedstock. (1)

(b) Regenerating the amine (stripping) requires large heat input — the energy penalty (~20–30% of a power plant's output). (1) If that energy comes from fossil fuels it emits $CO_2$ , partly offsetting the capture and reducing net benefit — a sustainability tension where capturing carbon costs energy that itself may carry a carbon footprint. (2)

[
  {"claim":"Route B ethylene oxide atom economy = 100%",
   "code":"M_C2H4=2*12.011+4*1.008; M_O=15.999; M_prod=M_C2H4+M_O; reactants=M_C2H4+0.5*2*M_O; ae=M_prod/reactants*100; result = abs(ae-100)<0.01"},
  {"claim":"Route A overall atom economy approx 32.4%",
   "code":"M_C2H4=2*12.011+4*1.008; M_Cl2=2*35.45; M_CaOH2=40.078+2*15.999+2*1.008; M_prod=M_C2H4+15.999; reactants=M_C2H4+M_Cl2+0.5*M_CaOH2; ae=M_prod/reactants*100; result = abs(ae-32.4)<0.3"},
  {"claim":"Electrolysis energy per mole water approx +237 kJ (n=2)",
   "code":"n=2; F=96485; E=-1.23; dG=-n*F*E; result = abs(dG/1000-237)<1"},
  {"claim":"Molar mass of ethylene oxide C2H4O equals reactant mass in route B",
   "code":"M_C2H4O=2*12.011+4*1.008+15.999; M_react=(2*12.011+4*1.008)+15.999; result = abs(M_C2H4O-M_react)<1e-9"}
]