Green Chemistry & Sustainability

Question 1

(a) Combined Route A (add step 1 ×2 + step 2, cancel intermediates): $2\,\text{C}_3\text{H}_6 + 2\,\text{Cl}_2 + 2\,\text{H}_2\text{O} + \text{Ca(OH)}_2 \rightarrow 2\,\text{C}_3\text{H}_6\text{O} + \text{CaCl}_2 + 2\,\text{HCl} + 2\,\text{H}_2\text{O}$ Net water: $2\text{H}_2\text{O}$ appears both sides → cancel: $2\,\text{C}_3\text{H}_6 + 2\,\text{Cl}_2 + \text{Ca(OH)}_2 \rightarrow 2\,\text{C}_3\text{H}_6\text{O} + \text{CaCl}_2 + 2\,\text{HCl}$ (2 marks balancing)

Atom economy = (mass desired product)/(mass all reactants)×100.

Reactant masses: C₃H₆=42.08 (×2=84.16); Cl₂=70.90 (×2=141.80); Ca(OH)₂=74.09. Sum = 300.05. (2) Desired product 2×58.08 = 116.16. (1) $AE_A = \frac{116.16}{300.05}\times100 = 38.7\%$ (1)

Route B: reactants C₃H₆=42.08 + H₂O₂=34.01 = 76.09; product 58.08. $AE_B = \frac{58.08}{76.09}\times100 = 76.3\%$ (2)

(b) (6, ~2 each; award for any three valid named principles applied)

• Principle 2 (Atom Economy): Route B much higher (76% vs 39%) — fewer atoms wasted.
• Principle 1 (Prevent waste): Route A produces CaCl₂ + HCl salt/acid waste streams; Route B's only byproduct is water — inherently less waste.
• Principle 3 (Less hazardous synthesis) / Principle 5 (Safer solvents): Route A uses Cl₂ (toxic, corrosive); Route B avoids chlorine.
• Principle 12 (Inherently safer chemistry for accident prevention): avoiding Cl₂ handling. Route B is greener. Why AE isn't the whole story: AE only counts atoms in the balanced equation; it ignores solvent, energy, catalyst toxicity, actual yield, and E-factor (waste per product). Route B still uses methanol solvent whose recovery/emissions matter → true greenness needs life-cycle/E-factor assessment, not just AE.

(c) (6) scCO₂ attractive because (2 each for two properties):

• Near critical point it has gas-like low viscosity & high diffusivity (fast mass transfer) yet liquid-like density/solvating power, tunable by pressure.
• Non-toxic, non-flammable, easily removed by depressurisation (no solvent residue), and recyclable/renewable (captured CO₂). Disadvantage (2): requires high-pressure equipment (p > 7.38 MPa) → capital/energy cost; poor solvent for polar/ionic species (unlike ionic liquids or water).

Question 2

(a) Reaction consumes 3 mol N₂H₄ per −336 kJ. Per mol: $-336/3 = -112\ \text{kJ mol}^{-1}$. (2) Moles per kg: $1000/32.05 = 31.20\ \text{mol}$. (1) Energy per kg: $31.20 \times 112 = 3494\ \text{kJ} \approx 3.49\ \text{MJ/kg}$ released. (2)

(b) (6) Advantages (any two, 2 each): (i) Much lower toxicity/vapour pressure than hydrazine (carcinogenic, high vapour toxicity) → cheaper handling, no SCAPE suits, faster ground ops. (ii) Higher density → more propellant in same tank (denser impulse). (iii) Reduced launch-site hazard/cost. Performance metric (2): Density specific impulse ($\rho I_{sp}$) — LMP-103S/AF-M315E typically ~6% higher $I_{sp}$ and notably higher density than hydrazine. (Accept "specific impulse $I_{sp}$" or "density impulse".)

(c) (9) $v_e = I_{sp}g_0$. Mass ratio: $m_0/m_f = e^{\Delta v/v_e}$, propellant $m_p = m_f(e^{\Delta v/v_e}-1)$.

Hydrazine: $v_e = 230\times9.81 = 2256.3\ \text{m/s}$; $\Delta v/v_e = 200/2256.3 = 0.08864$. $m_p = 500(e^{0.08864}-1) = 500(0.09269) = 46.34\ \text{kg}$. (3)

Green: $v_e = 252\times9.81 = 2472.1\ \text{m/s}$; $\Delta v/v_e = 200/2472.1 = 0.08090$. $m_p = 500(e^{0.08090}-1)=500(0.08427)=42.14\ \text{kg}$. (3)

Saving $= (46.34-42.14)/46.34\times100 = 9.06\%$. (3)

Question 3

(a) Voltage efficiency (electrolyser) $= E^\circ/V_{cell} = 1.23/1.80 = 0.683 = 68.3\%$. (2) Electrical energy per mol H₂: $W = nFV_{cell} = 2\times96485\times1.80 = 347{,}346\ \text{J} = 347.3\ \text{kJ}$. (4) (Reversible min would be $2\times96485\times1.23 = 237.4$ kJ = $-\Delta G$.)

(b) Fuel-cell efficiency $= V_{fc}/E^\circ = 0.70/1.23 = 0.569 = 56.9\%$. (2) Round-trip = (energy out)/(energy in) $= \dfrac{nFV_{fc}}{nFV_{el}} = \dfrac{V_{fc}}{V_{el}} = \dfrac{0.70}{1.80} = 0.389 = 38.9\%$. (3) (Equivalently product of the two efficiencies: $0.683\times0.569 = 0.389$.)

(c) (9: 5 code, 2 table values, 2 principle)

def roundtrip(V_el, V_fc, E0=1.23):
    eta_el = E0 / V_el        # electrolyser voltage efficiency
    eta_fc = V_fc / E0        # fuel cell voltage efficiency
    return eta_el * eta_fc    # = V_fc / V_el
 
for V in [1.6, 1.8, 2.0]:
    print(V, round(roundtrip(V, 0.70), 3))

Output:

Note round-trip = $V_{fc}/V_{el}$, independent of $E^\circ$. Principle: supports use of renewable feedstocks/energy (Principle 7) and energy efficiency / design for energy conservation (Principle 6) — H₂ stores surplus renewable electricity for later carbon-free power (hydrogen economy, decarbonisation).

[
  {"claim":"Route A atom economy ~38.7%","code":"AE=116.16/(84.16+141.80+74.09)*100\nresult = abs(AE-38.7)<0.3"},
  {"claim":"Route B atom economy ~76.3%","code":"AE=58.08/76.09*100\nresult = abs(AE-76.3)<0.3"},
  {"claim":"Hydrazine energy per kg ~3494 kJ","code":"E=(1000/32.05)*(336/3)\nresult = abs(E-3494)<10"},
  {"claim":"Propellant saving ~9.06%","code":"import math\nvh=230*9.81; vg=252*9.81\nmh=500*(math.exp(200/vh)-1); mg=500*(math.exp(200/vg)-1)\nsav=(mh-mg)/mh*100\nresult = abs(sav-9.06)<0.2"},
  {"claim":"Round-trip efficiency 1.8/0.70 = 0.389","code":"rt=0.70/1.80\nresult = abs(rt-0.389)<0.005"},
  {"claim":"Electrolyser energy per mol H2 ~347.3 kJ","code":"W=2*96485*1.80/1000\nresult = abs(W-347.3)<1"}
]