Question 4 of 6: Acid-Gas Removal, Isomerisation and Ethane Combustion

Question 4: Acid-Gas Removal, Isomerisation and Ethane Combustion (20 marks — equal value)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

(a) Three removal processes for each acid gas

Hydrogen sulphide (H₂S): (1) Amine absorption — chemical scrubbing with a regenerable alkanolamine (MEA, DEA, MDEA); H₂S reacts reversibly and is stripped in a regenerator, then routed to a Claus sulphur-recovery unit. (2) Caustic (alkaline) scrubbing — irreversible reaction with NaOH for smaller streams. (3) Solid-bed adsorption / scavenging — iron sponge (iron oxide) or zinc oxide beds, $ZnO + H_2S \rightarrow ZnS + H_2O$, giving very deep polishing.

Carbon dioxide (CO₂): (1) Amine absorption — the same regenerable alkanolamine circuit (MDEA is often selected for CO₂). (2) Physical-solvent absorption — Selexol or Rectisol (chilled methanol), which dissolve CO₂ physically and are favoured at high partial pressures. (3) Membrane separation — polymeric membranes that preferentially permeate CO₂; molecular-sieve adsorption is a further option for final drying/polishing.

(b) Importance of isomerisation

Isomerisation rearranges straight-chain (normal) paraffins into their branched isomers without changing molecular weight, e.g. $n\text{-}C_5/n\text{-}C_6 \rightarrow$ iso-pentane / iso-hexanes over a Pt / chlorided-alumina or zeolite catalyst at low temperature. Its importance is twofold: (i) it markedly raises the octane number of the light naphtha (branched paraffins knock far less than n-paraffins) without adding aromatics such as benzene, helping refiners meet clean-gasoline specifications; and (ii) it converts n-butane to isobutane, the essential feedstock for alkylation. It is a low-severity, high-value route to premium gasoline blendstock.

(c) Flue-gas composition from ethane combustion

Given. Feed gas (basis 100 mol): 80 mol ethane (C₂H₆) and 20 mol O₂, burned with 200% excess air. Of the ethane, 80% → CO₂, 10% → CO, 10% remains unburned.

Find. The molar composition of the stack (flue) gas.

Approach. Establish the theoretical oxygen for complete combustion (crediting the O₂ already in the feed), size the air from the 200% excess, run the actual (partial) reactions to get products and oxygen consumed, then tally every stack component.

1. Theoretical oxygen. Complete combustion is $C_2H_6+\tfrac{7}{2}O_2\rightarrow 2CO_2+3H_2O$, so burning all 80 mol ethane needs $80\times3.5=280$ mol O₂. The feed already supplies 20 mol O₂, so the air must furnish a net theoretical $$O_{2,\text{theo}}^{\text{air}} = 280-20 = 260\ \text{mol}.$$
2. Air supplied at 200% excess. "200% excess" means the air O₂ is three times theoretical: $$O_{2,\text{air}} = 3\times260 = 780\ \text{mol},\qquad N_2 = 780\times\frac{79}{21} = 2934.3\ \text{mol}.$$
3. Actual reactions. $64$ mol ethane (80%) burns to CO₂, $8$ mol (10%) to CO, $8$ mol (10%) is unburned: $$64\,C_2H_6 \rightarrow 128\,CO_2 + 192\,H_2O\ \ (\text{O}_2\ \text{used}=224),$$ $$8\,C_2H_6 \rightarrow 16\,CO + 24\,H_2O\ \ (\text{O}_2\ \text{used}=20).$$ Total O₂ consumed $=224+20=244$ mol.
4. Oxygen leaving. Oxygen in $=$ air + feed $=780+20=800$ mol, so $$O_{2,\text{out}} = 800-244 = 556\ \text{mol}.$$
5. Assemble the stack gas. Collecting all species (H₂O totals $192+24=216$ mol): $$CO_2=128,\ CO=16,\ H_2O=216,\ O_2=556,\ N_2=2934.3,\ C_2H_6=8,$$ $$n_{\text{total}} = \boxed{3858.3\ \text{mol}}.$$
6. Mole percents. Divide each by the total: $$\boxed{CO_2\ 3.32\%,\ CO\ 0.41\%,\ H_2O\ 5.60\%,\ O_2\ 14.41\%,\ N_2\ 76.05\%,\ C_2H_6\ 0.21\%.}$$