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Chapter 8 β€’ Theory & Derivations

Unit 8: Chlor-Alkali and Heavy Chemicals: Membrane Cells & Heavy Acids

Exhaustive electrochemical and process engineering treatise on the chlor-alkali industry: primary brine precipitation, secondary chelating ion-exchange resin purification, membrane cell electrochemistry (Nafion perfluorosulfonate membranes, DSA anodes, nickel gas-diffusion cathodes), cell voltage decomposition (overpotentials and ohmic drops), caustic concentration, chlorine liquefaction, and industrial sulfuric acid manufacture via the DCDA contact process.

Β§8.1 Brine Chemistry & Ultra-Pure Purification: Primary Treatment & Chelating Resins

The raw feedstock for the chlor-alkali industry is saturated aqueous sodium chloride ($\text{NaCl}$) brine ($300 - 320\text{ g/L NaCl}$) sourced from solution-mined underground salt deposits or solar marine salt pans.

Impurity Hazards in Modern Membrane Cells

Modern perfluorinated cation-exchange membranes (e.g., DuPont Nafion, Asahi Kasei Aciplex) are irreversibly poisoned by multivalent alkaline-earth cations ($\text{Ca}^{2+}, \text{Mg}^{2+}, \text{Ba}^{2+}, \text{Sr}^{2+}$). Inside the membrane, migrating divalent cations encounter hydroxide ions ($\text{OH}^-$) diffusing from the catholyte:

$$\text{Ca}^{2+} + 2\text{OH}^- \longrightarrow \text{Ca(OH)}_2(s) \downarrow$$
$$\text{Mg}^{2+} + 2\text{OH}^- \longrightarrow \text{Mg(OH)}_2(s) \downarrow$$

Insoluble precipitates crystallize inside the microscopic hydrophilic ion clusters ($2 - 4\text{ nm}$), rupturing polymer chains, causing blistering, and escalating electrical cell voltage. Consequently, total hardness ($[\text{Ca}^{2+}] + [\text{Mg}^{2+}]$) must be reduced below $20\text{ ppb}$ ($0.020\text{ mg/L}$).

Two-Stage Industrial Purification Circuit

1. Primary Chemical Precipitation:

  • Sodium carbonate ($\text{Na}_2\text{CO}_3$) precipitates calcium:
$$\text{Ca}^{2+} + \text{CO}_3^{2-} \longrightarrow \text{CaCO}_3(s) \downarrow$$
  • Sodium hydroxide ($\text{NaOH}$) precipitates magnesium and iron:
$$\text{Mg}^{2+} + 2\text{OH}^- \longrightarrow \text{Mg(OH)}_2(s) \downarrow$$
  • Barium chloride ($\text{BaCl}_2$) precipitates sulfate:
$$\text{SO}_4^{2-} + \text{Ba}^{2+} \longrightarrow \text{BaSO}_4(s) \downarrow$$

The effluent is treated with polyelectrolyte flocculants, clarified in rake settlers, and polished through anthracite/sand filters, reducing hardness to $1 - 5\text{ ppm}$.

2. Secondary Purification via Chelating Ion-Exchange Resins:

The polished brine passes through packed columns of macroporous polystyrene resins functionalized with iminodiacetic acid or aminomethylphosphonic acid groups:

$$2\,\text{R-CH}_2\text{-N(CH}_2\text{COO}^-\text{Na}^+)_2 + \text{Ca}^{2+} \rightleftharpoons [\text{R-CH}_2\text{-N(CH}_2\text{COO}^-)_2]_2\text{Ca}^{2+} + 2\text{Na}^+$$

The chelating resin exhibits an affinity for divalent alkaline earths over sodium ions exceeding $10,000 : 1$, consistently discharging ultra-pure brine containing $< 10\text{ ppb total hardness}$.

Β§8.2 Chlor-Alkali Technologies: Mercury, Diaphragm & Modern Cation-Exchange Membrane Cells

The industrial electrolysis of aqueous sodium chloride has undergone three major technological transitions:

Comparison of Electrolytic Technologies

| Feature | Mercury Cell (Castner-Kellner) | Diaphragm Cell | Modern Membrane Cell | |---|---|---|---| | Anode Reaction | $2\text{Cl}^- \to \text{Cl}_2 + 2e^-$ | $2\text{Cl}^- \to \text{Cl}_2 + 2e^-$ | $2\text{Cl}^- \to \text{Cl}_2 + 2e^-$ | | Cathode Reaction | $\text{Na}^+ + \text{Hg} + e^- \to \text{Na(Hg)}$ amalgam | $2\text{H}_2\text{O} + 2e^- \to \text{H}_2 + 2\text{OH}^-$ | $2\text{H}_2\text{O} + 2e^- \to \text{H}_2 + 2\text{OH}^-$ | | Caustic Purity | Pure $50\%\text{ NaOH}$ directly (from decomposer) | Dilute $12\%\text{ NaOH} + 15\%\text{ NaCl}$ (requires huge evaporation) | Pure $32 - 35\%\text{ NaOH}$ ($< 30\text{ ppm NaCl}$) | | Electrical Energy | $3,100 - 3,400\text{ kWh/t NaOH}$ | $2,700 - 3,000\text{ kWh/t NaOH}$ | $2,100 - 2,400\text{ kWh/t NaOH}$ | | Environmental Hazard | Severe toxic mercury bioaccumulation (Minamata) | Carcinogenic asbestos fiber emission | Completely environmentally benign |

``` MODERN MEMBRANE ELECTROLYSIS CELL Depleted Brine (200 g/L) Water / Dilute NaOH (30%) β–² β–² β”‚ β”‚ β”Œβ”€β”€β”€β”€β”΄β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”Œβ”€β”€β”€β”€β”€β”€β”΄β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ ANODE COMPARTMENT β”‚ β”‚ CATHODE COMPARTMENT β”‚ β”‚ β”‚ β”‚ β”‚ Saturated ──> [ DSA Ti Mesh Anode]β”‚ β”‚[ Nickel Mesh Cathode ] <── Dilute Brine β”‚ 2 Cl⁻ ──> Clβ‚‚ + 2e⁻ β”‚ β”‚ 2 Hβ‚‚O + 2e⁻ ──> Hβ‚‚ + 2OH⁻ β”‚ NaOH Feed (300 g/L) β”‚ β”‚ β”‚ β”‚ β”‚ β”‚ β”‚ β”‚ Clβ‚‚ Gas ──> β”‚ Na⁺ β”‚ <── Hβ‚‚ Gas β”‚ β”‚ β”‚ ──────>β”‚ β”‚ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β”‚ Nafion Perfluorinated β”‚ β”‚ Cation-Exchange Membrane β–Ό β”‚ (Sulfonate / Carboxylate Layers) Pure Product NaOH (32-35%) ```

Membrane Architecture (Bilayer Design)

Modern membranes consist of a reinforced perfluorosulfonic acid (PFSA) backing layer facing the anode (low electrical resistance) coupled to an ultra-thin carboxylate polymer layer facing the cathode. The high fixed charge density of carboxylate groups ($\text{-COO}^-$) creates extreme Donnan exclusion against back-migrating hydroxide ions ($\text{OH}^-$), maintaining caustic current efficiencies above $95 - 97\%$ at $35\text{ wt}\%\text{ NaOH}$.

Β§8.3 Membrane Cell Electrochemistry: Cell Potential, Overpotentials & Ohmic Drops

The overall electrochemical decomposition of aqueous sodium chloride:

$$2\text{NaCl}(aq) + 2\text{H}_2\text{O}(l) \longrightarrow 2\text{NaOH}(aq) + \text{Cl}_2(g) + \text{H}_2(g)$$

Thermochemically, $\Delta G^\circ = +422.3\text{ kJ/mol}$. The reversible thermodynamic cell potential ($E_{\text{rev}}^\circ$) under standard conditions ($298\text{ K}$, $1\text{ bar}$):

$$E_{\text{rev}}^\circ = - \frac{\Delta G^\circ}{n F} = - \frac{422,300\text{ J}}{2 \times 96,485\text{ C}} = -2.188\text{ V}$$

Cell Voltage Breakdown ($U_{\text{cell}}$)

The actual operational cell voltage required to drive electrolysis at commercial current densities ($j = 4.0 - 7.0\text{ kA/m}^2$) is significantly higher, expressed by the additive polarization sum:

$$U_{\text{cell}} = |E_{\text{rev}}| + \eta_{\text{anode}} + |\eta_{\text{cathode}}| + I \cdot R_{\text{membrane}} + I \cdot R_{\text{electrolyte}} + I \cdot R_{\text{structure}}$$

1. Reversible Potential under Operating Conditions ($90^\circ\text{C}$):

$$E_{\text{rev}}(90^\circ\text{C}) \approx 2.15\text{ V}$$

2. Anodic Overpotential ($\eta_{\text{anode}}$):

Chlorine evolution on modern Dimensionally Stable Anodes (DSA: titanium substrate coated with mixed metal oxides $\text{RuO}_2-\text{TiO}_2-\text{IrO}_2$). Catalytic activation lowers overpotential to $\eta_{\text{anode}} \approx 0.05 - 0.08\text{ V}$ via the Butler-Volmer relation:

$$\eta_a = \frac{RT}{\alpha_a F} \ln\left(\frac{j}{j_{0,a}}\right)$$

3. Cathodic Overpotential ($\eta_{\text{cathode}}$):

Hydrogen evolution on activated nickel cathodes coated with ruthenium or Raney nickel. $\eta_{\text{cathode}} \approx 0.08 - 0.12\text{ V}$.

4. Ohmic Voltage Drop Across Membrane ($I \cdot R_{\text{membrane}}$):

Resistance of $\text{Na}^+$ transport across the membrane: $\Delta U_{\text{mem}} \approx 0.35 - 0.50\text{ V}$.

5. Ohmic Drops in Electrolytes and Structural Hardware:

Bubble void fraction (gas dispersion of $\text{Cl}_2$ and $\text{H}_2$) elevates solution resistance by the Bruggeman relation:

$$R = R_0 (1 - \epsilon)^{-1.5}$$

Zero-gap cell configurations (where flexible mesh electrodes compress directly against the membrane) compress inter-electrode electrolyte drops to $< 0.10\text{ V}$. Total industrial operational cell voltage sits between $2.95\text{ V}$ and $3.15\text{ V}$ at $6.0\text{ kA/m}^2$.

Complete Electrochemical Parameters of Modern Chlor-Alkali Membrane Cells

| Electrochemical Parameter | Symbol | Industrial Operating Range | Reference Target Value | |---|---|---|---| | Current Density | $j$ | $4.0 - 7.0\text{ kA/m}^2$ | $6.0\text{ kA/m}^2$ | | Operating Temperature | $T$ | $85 - 92^\circ\text{C}$ | $88^\circ\text{C}$ | | Anolyte NaCl Concentration | $[\text{NaCl}]_{\text{ano}}$ | $190 - 220\text{ g/L}$ | $205\text{ g/L}$ | | Catholyte NaOH Concentration| $[\text{NaOH}]_{\text{cat}}$ | $32.0 - 35.0\text{ wt}\%$ | $33.5\text{ wt}\%$ | | Current Efficiency (NaOH) | $\eta_{\text{NaOH}}$ | $95.5 - 97.5\%$ | $96.5\%$ | | Operating Cell Voltage | $U_{\text{cell}}$ | $2.95 - 3.15\text{ V}$ | $3.02\text{ V}$ | | Specific Power Consumption | $w_{\text{spec}}$ | $2,050 - 2,250\text{ kWh/t NaOH}$ | $2,120\text{ kWh/t NaOH}$ | | Membrane Service Life | $\tau_{\text{mem}}$ | $3 - 5\text{ years}$ | $4\text{ years}$ | | Anode Coating Service Life | $\tau_{\text{anode}}$ | $8 - 12\text{ years}$ | $10\text{ years}$ |

Β§8.4 Caustic Soda Processing: Multi-Effect Evaporative Concentration & Flaking

The aqueous sodium hydroxide discharged from modern membrane cells has a concentration of $32 - 35\text{ wt}\%\text{ NaOH}$. Standard global merchant commerce requires $50.0\text{ wt}\%\text{ NaOH}$ liquid caustic soda or $99\%\text{ NaOH}$ anhydrous solid pearls/flakes.

1. Multi-Effect Falling-Film Evaporation ($32\% \to 50\%\text{ NaOH}$)

Concentrating caustic soda requires specialized metallurgy (pure nickel $\text{Ni } 200$ or high-nickel alloys) to resist caustic stress-corrosion cracking and embrittlement at elevated temperatures ($> 120^\circ\text{C}$):

  • Modern plants employ triple- or quadruple-effect falling film evaporators operating under forward or counter-current feed.
  • Boiling Point Elevation (BPE): Saturated caustic solutions exhibit extreme boiling point elevation. While water boils at $100^\circ\text{C}$ at $1\text{ atm}$, a $50\text{ wt}\%\text{ NaOH}$ solution boils at $143^\circ\text{C}$ ($\text{BPE} = 43\text{ K}$). At $73\text{ wt}\%\text{ NaOH}$, the boiling point climbs to $190^\circ\text{C}$. This massive BPE significantly compresses the effective logarithmic mean temperature difference ($\Delta T_{\text{eff}}$) available across successive evaporator effects.

2. Solid Anhydrous Caustic Production ($50\% \to 99\%\text{ NaOH}$)

To produce solid flake or pearl caustic:

  • The $50\text{ wt}\%$ caustic is concentrated in a falling film evaporator heated by molten heat transfer salts (eutectic mixture of $53\%\text{ KNO}_3 + 40\%\text{ NaNO}_2 + 7\%\text{ NaNO}_3$) operating at $380 - 400^\circ\text{C}$ under vacuum.
  • Molten anhydrous caustic leaves at $320 - 340^\circ\text{C}$ ($< 0.5\%\text{ moisture}$).
  • Flaking: The molten liquid is fed onto water-cooled rotating nickel flaker drums, solidifying instantaneously into a crystalline sheet that is sheared off by doctor blades.
  • Prilling: Alternatively, molten caustic is sprayed down a counter-current air prilling tower, crystallizing into spherical beads (pearls).

Β§8.5 Chlorine Gas Engineering: Cooling, Sulfuric Acid Drying, Compression & Liquefaction

Moist, saturated chlorine gas ($\text{Cl}_2$) discharged from the membrane cell anodes at $85 - 90^\circ\text{C}$ is saturated with water vapor and is intensely corrosive to all conventional structural metals:

``` CHLORINE GAS PROCESSING TRAIN Wet Hot Cl2 Gas (~85Β°C) from Cells β”‚ β–Ό β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ DIRECT WATER β”‚ (Chilled water spray cools Cl2 to 12-15Β°C; β”‚ COOLER β”‚ condenses >80% of water vapor without hydrate formation) β””β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”˜ β–Ό Cooled Wet Cl2 β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ SULFURIC ACID DRYING TOWERS (3-Stage Counter-Current) β”‚ <── Fresh 98% H2SO4 β”‚ Stage 1 (78% H2SO4) ──> Stage 2 (92%) ──> Stage 3 (96-98%)β”‚ ──> Spent Dilute H2SO4 β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β–Ό Bone-Dry Cl2 Gas (Moisture < 5 ppm) β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ CHLORINE COMPRESSOR (Liquid Ring or Centrifugal) β”‚ ──> Pressurizes to 8-12 bar β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β–Ό β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ LIQUEFACTION CONDENSER (Refrigerated Freon/Ammonia) β”‚ ──> Pure Liquid Chlorine (-15Β°C) β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”¬β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ β–Ό β”Œβ”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β” β”‚ STORAGE TANK β”‚ ──> Bulk Pressurized Railcars / Pipeline Distribution β””β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”€β”˜ ```

1. Direct Gas Cooling & Hydrate Prevention

The gas is cooled to $12 - 15^\circ\text{C}$ in packed titanium or polyvinylidene fluoride (PVDF) direct-contact cooling towers using chilled water. Cooling must strictly remain above $9.8^\circ\text{C}$ at atmospheric pressure to prevent crystallization of solid yellow chlorine hydrate ($\text{Cl}_2\cdot 7.3\text{H}_2\text{O}$), which clogs tower packing and transfer pipelines.

2. Multi-Stage Sulfuric Acid Drying

Water must be removed to $< 5\text{ ppm}$ moisture:

  • When moisture $< 20\text{ ppm}$, chlorine is non-corrosive to ordinary carbon steel, allowing inexpensive steel pipes, valves, and railcars to be used safely.
  • Drying is conducted in three packed towers flowing concentrated sulfuric acid ($\text{H}_2\text{SO}_4$) counter-currently. Fresh $98\text{ wt}\%\text{ H}_2\text{SO}_4$ enters the third stage, cascading to the first stage where spent acid exits at $75 - 78\text{ wt}\%$.

3. Compression & Liquefaction

Bone-dry chlorine is compressed using liquid-ring compressors (employing concentrated sulfuric acid as the sealing fluid) or multi-stage centrifugal compressors to $8 - 12\text{ bar}$. The pressurized gas enters shell-and-tube condensers chilled by refrigerant (ammonia or R-134a) to $-10^\circ\text{C}$ to $-25^\circ\text{C}$, condensing into clear amber liquid chlorine ($\rho = 1.41\text{ g/cm}^3$). Non-condensable inert gases ($\text{H}_2, \text{N}_2, \text{O}_2$) are purged ("sniff gas") to a sodium hydroxide scrubber.

Β§8.6 Hydrochloric Acid & Sodium Hypochlorite Synthesis Engineering

Byproduct chlorine and hydrogen from the chlor-alkali cell room are converted on-site into essential industrial chemicals:

1. Hydrochloric Acid ($\text{HCl}$) Synthesis

Pure gaseous hydrogen and chlorine are reacted in a specialized water-cooled silica or impregnated graphite combustion chamber:

$$\text{H}_2(g) + \text{Cl}_2(g) \longrightarrow 2\text{HCl}(g) \quad \Delta H_{298}^\circ = -184.6\text{ kJ/mol} \quad (-92.3\text{ kJ/mol HCl})$$
  • Combustion Control: To ensure complete consumption of chlorine (which is intensely toxic and corrosive), hydrogen is fed at a $5 - 10\%$ stoichiometric excess. The flame temperature exceeds $2000 - 2500^\circ\text{C}$.
  • Adiabatic Absorption: The exiting hot anhydrous $\text{HCl}$ gas enters an isothermal or adiabatic falling-film graphite absorption tower where it dissolves violently in demineralized water:
$$\text{HCl}(g) + n\text{H}_2\text{O}(l) \longrightarrow \text{HCl}(aq) \quad (\Delta H_{\text{abs}} = -74.8\text{ kJ/mol})$$

The maximum concentration at atmospheric pressure is governed by the negative azeotrope ($20.22\text{ wt}\%\text{ HCl}$ at $108.6^\circ\text{C}$), but refrigerated commercial absorbers produce concentrated $33 - 36\text{ wt}\%\text{ Technical Grade HCl}$.

2. Sodium Hypochlorite ($\text{NaOCl}$) Bleach Manufacture

Produced by scrubbing dilute or tail-gas chlorine with refrigerated aqueous sodium hydroxide:

$$\text{Cl}_2(g) + 2\text{NaOH}(aq) \longrightarrow \text{NaOCl}(aq) + \text{NaCl}(aq) + \text{H}_2\text{O}(l) \quad \Delta H^\circ = -103\text{ kJ/mol}$$
  • Temperature & Decomposition Control: The reaction is conducted below $30 - 35^\circ\text{C}$. Above $40^\circ\text{C}$, hypochlorite decomposes rapidly into toxic and inactive chlorate:
$$3\text{NaOCl} \overset{\Delta}{\longrightarrow} 2\text{NaCl} + \text{NaClO}_3$$
  • Free Alkali Stabilization: Excess sodium hydroxide ($0.5 - 1.0\text{ wt}\%\text{ free NaOH}$) is maintained to keep the $\text{pH} > 11.5$, preventing decomposition into hypochlorous acid ($\text{HOCl}$) and toxic chlorine gas.

Β§8.7 Sulfuric Acid Manufacture: The Modern Double Contact Double Absorption (DCDA) Process

Sulfuric acid ($\text{H}_2\text{SO}_4$, $M = 98.08\text{ g/mol}$) is the world's most widely consumed heavy industrial chemical. Modern production utilizes the Double Contact Double Absorption (DCDA) catalytic contact process:

1. Sulfur Combustion & Gas Conditioning

Molten bright sulfur ($135 - 145^\circ\text{C}$) is atomized with dry air in a refractory burner:

$$\text{S}(l) + \text{O}_2(g) \longrightarrow \text{SO}_2(g) \quad \Delta H = -296.8\text{ kJ/mol}$$

The resulting process gas ($10 - 11.5\text{ vol}\%\text{ SO}_2$, $9.5 - 11\text{ vol}\%\text{ O}_2$) is cooled from $1050^\circ\text{C}$ to $420^\circ\text{C}$ in a waste heat boiler, generating high-pressure steam ($40 - 60\text{ bar}$).

2. Catalytic Oxidation Kinetics ($SO_2 \to SO_3$)

The reversible oxidation is conducted over a cesium-promoted vanadium pentoxide catalyst supported on silica ($\text{V}_2\text{O}_5-\text{K}_2\text{SO}_4/\text{SiO}_2$):

$$\text{SO}_2(g) + \frac{1}{2}\text{O}_2(g) \rightleftharpoons \text{SO}_3(g) \quad \Delta H_{298}^\circ = -98.9\text{ kJ/mol}$$

The equilibrium constant is given by:

$$\log_{10} K_p = \frac{5,005}{T} - 4.743 \quad (T\text{ in K})$$

Because the reaction is strongly exothermic, thermodynamic equilibrium conversion decreases with increasing temperature, while reaction kinetics freeze below $400^\circ\text{C}$. The converter is structured into four sequential adiabatic catalyst beds with inter-bed cooling:

  • Bed 1 ($420 \to 600^\circ\text{C}$): Rapid kinetic conversion reaches $60 - 65\%$.
  • Bed 2 ($440 \to 510^\circ\text{C}$): Cumulative conversion reaches $85\%$.
  • Bed 3 ($430 \to 455^\circ\text{C}$): Conversion reaches $93 - 95\%$.

3. The DCDA Innovation (Interpass Absorption)

In a single absorption plant, Le Chatelier's principle restricts overall conversion to $\le 97.5\%$, discharging thousands of ppm of harmful $\text{SO}_2$ into the atmosphere. In the DCDA configuration:

  • Process gas leaving Bed 3 is cooled and passed through an Intermediate Absorption Tower (IPAT) where $> 99.9\%$ of generated $\text{SO}_3$ is absorbed into circulating $98.5\%\text{ H}_2\text{SO}_4$.
  • The $\text{SO}_3$-depleted gas is reheated to $420^\circ\text{C}$ in gas-gas heat exchangers and fed to Bed 4.
  • Removing $\text{SO}_3$ drives the thermodynamic equilibrium of the remaining gas overwhelmingly forward, raising overall conversion to $> 99.85\%$, cutting stack emissions to $< 100\text{ ppm SO}_2$.
  • The gas exits Bed 4 through a Final Absorption Tower (FAT), producing concentrated $98.5\text{ wt}\%\text{ H}_2\text{SO}_4$ and Oleum ($20 - 65\%\text{ free SO}_3$).

Β§8.8 Oxygen-Depolarized Cathodes (ODC), Green Hydrogen & Zero-Emission Chlor-Alkali

The chlor-alkali sector consumes massive electrical energy ($> 2,200\text{ kWh/metric ton NaOH}$). The development of Oxygen-Depolarized Cathode (ODC) technology and hydrogen valorization represents a paradigm shift toward decarbonized heavy chemicals:

1. Oxygen-Depolarized Cathode (ODC) Electrochemistry

In conventional membrane cells, hydrogen gas is evolved at the cathode by water reduction:

$$2\text{H}_2\text{O} + 2e^- \longrightarrow \text{H}_2(g) + 2\text{OH}^- \quad (E^\circ = -0.828\text{ V})$$

In an ODC cell, gaseous pure oxygen ($\text{O}_2$) is introduced through a porous gas-diffusion cathode coated with silver or platinum catalysts, depolarizing the cathode reaction into oxygen reduction:

$$\text{O}_2(g) + 2\text{H}_2\text{O} + 4e^- \longrightarrow 4\text{OH}^- \quad (E^\circ = +0.401\text{ V})$$
  • Thermodynamic Voltage Reduction:

The cathode standard potential shifts positively by $\Delta E^\circ = 0.401 - (-0.828) = +1.229\text{ V}$.

  • Cell Operating Voltage Drop:

Operational cell voltage plummets from $\sim 3.00\text{ V}$ to $2.00 - 2.10\text{ V}$ at $4 - 5\text{ kA/m}^2$.

  • Energy Conservation:

Electrical energy consumption drops from $2,200\text{ kWh/t NaOH}$ to $1,500 - 1,600\text{ kWh/t NaOH}$, achieving an extraordinary $30\%$ reduction in grid power consumption.

2. High-Purity Green Hydrogen Valorization

For conventional membrane plants that continue to produce byproduct hydrogen:

  • Chlor-alkali hydrogen is ultra-pure ($> 99.999\%$ after moisture condensation and trace oxygen catalytic deoxidation).
  • Rather than combusting hydrogen for low-grade process steam, modern complexes feed this chemical-grade hydrogen into fuel-cell vehicles, direct ammonia synthesis, or green methanol plants, capturing immense clean-energy carbon credits.

University Honors Industrial Case Study: Wet Chlorine Crevice Corrosion in Titanium Exchangers

Titanium is the universal metal of choice for handling wet chlorine gas because a stable, self-healing rutile passive oxide film ($\text{TiO}_2$) forms instantaneously in the presence of trace moisture ($> 0.5\text{ wt}\%\text{ H}_2\text{O}$):

$$\text{Ti} + 2\text{H}_2\text{O} \longrightarrow \text{TiO}_2 + 4\text{H}^+ + 4e^-$$
  • The Dry Chlorine Fire Hazard: In bone-dry chlorine gas ($< 20\text{ ppm H}_2\text{O}$), the protective oxide film cannot self-repair. If mechanical scratching exposes virgin titanium metal, a violent, auto-igniting exothermic chlorination fire erupts:
$$\text{Ti}(s) + 2\text{Cl}_2(g) \longrightarrow \text{TiCl}_4(l) \quad (\Delta H = -804\text{ kJ/mol})$$

The titanium metal burns vigorously in dry chlorine at room temperature, releasing dense white clouds of boiling $\text{TiCl}_4$.

  • Crevice Corrosion: In gasket joints of plate heat exchangers where brine flow stagnates and temperature exceeds $75^\circ\text{C}$, localized acid buildup ($\text{pH} < 1$) breaks down passivity. Modern chlor-alkali exchangers specify palladium-stabilized titanium (ASTM Grade 7, $\text{Ti}-0.15\%\text{Pd}$) to prevent crevice initiation.
Easy Example 8.1: Faraday's Law & Chlor-Alkali Membrane Cell Yields

A chlor-alkali plant operates an electrolyzer circuit containing $120$ membrane cells connected in electrical series.

  • Operational direct current is $I = 15,000\text{ A}$ ($15.0\text{ kA}$).
  • Current efficiency for sodium hydroxide ($\text{NaOH}$, $40.00\text{ g/mol}$) is $\eta_{\text{NaOH}} = 96.0\%$.
  • Current efficiency for chlorine gas ($\text{Cl}_2$, $70.90\text{ g/mol}$) is $\eta_{\text{Cl}_2} = 94.5\%$.
  • Current efficiency for hydrogen gas ($\text{H}_2$, $2.016\text{ g/mol}$) is $\eta_{\text{H}_2} = 99.0\%$.

Faraday's constant is $F = 96,485\text{ C/mol}$.

  1. Calculate the daily production rate of pure $100\%\text{ NaOH}$ in metric tons per day.
  2. Determine the hourly mass and STP volumetric flow rate of chlorine gas ($\text{Cl}_2$) in $\text{kg/h}$ and $\text{Nm}^3\text{/h}$ ($22.414\text{ Nm}^3\text{/kmol}$).
  3. Calculate the hourly mass and STP volumetric flow rate of hydrogen gas in $\text{kg/h}$ and $\text{Nm}^3\text{/h}$.

Step 1: Daily $\text{NaOH}$ Production Rate

Since all $N = 120$ cells are in series, the same current $I$ passes through each cell. Total charge passed across the entire circuit per day ($t = 24\text{ h} = 86,400\text{ s}$):

$$Q_{\text{day}} = N \times I \times t = 120 \times 15,000\text{ A} \times 86,400\text{ s} = 1.5552 \times 10^{11}\text{ Coulombs}$$

Moles of electrons transferred per day:

$$n_e = \frac{Q_{\text{day}}}{F} = \frac{1.5552 \times 10^{11}\text{ C}}{96,485\text{ C/mol}} = 1,611,857\text{ mol } e^- = 1,611.857\text{ kmol } e^-$$

Since $1\text{ mol } e^-$ produces $1\text{ mol NaOH}$:

$$n_{\text{NaOH}} = \eta_{\text{NaOH}} \times n_e = 0.960 \times 1,611.857 = 1,547.38\text{ kmol/day}$$

Mass of pure $\text{NaOH}$:

$$m_{\text{NaOH}} = 1,547.38\text{ kmol} \times 40.00\text{ kg/kmol} = 61,895\text{ kg/day} \approx 61.90\text{ metric tons/day}$$

Step 2: Chlorine Production Rates

Total charge passed per hour:

$$Q_{\text{hour}} = 120 \times 15,000\text{ A} \times 3,600\text{ s} = 6.48 \times 10^9\text{ C}$$

Moles of electrons per hour:

$$n_{e, \text{hour}} = \frac{6.48 \times 10^9\text{ C}}{96,485\text{ C/mol}} = 67,160.7\text{ mol } e^-/\text{h} = 67.161\text{ kmol } e^-/\text{h}$$

Chlorine generation requires $2\text{ mol } e^-$ per mole of $\text{Cl}_2$:

$$n_{\text{Cl}_2} = \eta_{\text{Cl}_2} \times \frac{n_{e, \text{hour}}}{2} = 0.945 \times \frac{67.161}{2} = 31.734\text{ kmol/h}$$

Mass of chlorine per hour:

$$\dot{m}_{\text{Cl}_2} = 31.734\text{ kmol/h} \times 70.90\text{ kg/kmol} = 2,249.9\text{ kg/h} \approx 2.250\text{ metric tons/h}$$

Volumetric flow rate at STP:

$$\dot{V}_{\text{Cl}_2, \text{STP}} = 31.734\text{ kmol/h} \times 22.414\text{ Nm}^3\text{/kmol} = 711.29\text{ Nm}^3\text{/h}$$

Step 3: Hydrogen Production Rates

Hydrogen generation requires $2\text{ mol } e^-$ per mole of $\text{H}_2$:

$$n_{\text{H}_2} = \eta_{\text{H}_2} \times \frac{n_{e, \text{hour}}}{2} = 0.990 \times \frac{67.161}{2} = 33.245\text{ kmol/h}$$

Mass of hydrogen:

$$\dot{m}_{\text{H}_2} = 33.245\text{ kmol/h} \times 2.016\text{ kg/kmol} = 67.02\text{ kg/h}$$

Volumetric flow rate at STP:

$$\dot{V}_{\text{H}_2, \text{STP}} = 33.245\text{ kmol/h} \times 22.414\text{ Nm}^3\text{/kmol} = 745.15\text{ Nm}^3\text{/h}$$

The plant produces $61.9\text{ t/day}$ NaOH, $2.25\text{ t/h}$ ($711.3\text{ Nm}^3\text{/h}$) Cl2, and $67.0\text{ kg/h}$ ($745.2\text{ Nm}^3\text{/h}$) H2.

Medium Example 8.2: Membrane Cell Voltage Decomposition & Specific Power Consumption

A membrane electrolyzer operates at a current density of $j = 6.00\text{ kA/m}^2$ ($6,000\text{ A/m}^2$) at $90^\circ\text{C}$. The components of the cell voltage are measured as:

  • Reversible thermodynamic cell potential: $E_{\text{rev}} = 2.160\text{ V}$
  • Anodic chlorine overpotential: $\eta_a = 0.065\text{ V}$
  • Cathodic hydrogen overpotential: $|\eta_c| = 0.095\text{ V}$
  • Area-specific membrane resistance: $r_{\text{mem}} = 0.060\text{ \Omega}\cdot\text{m}^2$
  • Area-specific electrolyte solution resistance: $r_{\text{sol}} = 0.025\text{ \Omega}\cdot\text{m}^2$
  • Structural hardware and contact resistance: $\Delta U_{\text{struct}} = 0.040\text{ V}$

Current efficiency for $\text{NaOH}$ ($40.00\text{ g/mol}$) is $\eta = 96.5\%$. Faraday's constant is $F = 96,485\text{ C/mol}$.

  1. Calculate the total operational cell voltage ($U_{\text{cell}}$) in volts.
  2. Determine the fraction of electrical voltage consumed by thermodynamic work versus irreversible overpotentials and ohmic dissipation.
  3. Calculate the specific direct-current electrical energy consumption per metric ton of pure $\text{NaOH}$ in $\text{kWh/t NaOH}$.

Step 1: Total Cell Voltage Calculation

Calculate the ohmic voltage drops from current density ($j = 6,000\text{ A/m}^2$):

$$\Delta U_{\text{mem}} = j \cdot r_{\text{mem}} = 6,000\text{ A/m}^2 \times 0.060 \times 10^{-4}\text{ \Omega}\cdot\text{m}^2 \text{ (note: standard area resistance is }\sim 0.6 \times 10^{-4}\text{ \Omega}\cdot\text{m}^2\text{)}$$

Using consistent units with $j \cdot r_{\text{mem}} = 6.00\text{ kA/m}^2 \times 0.060\text{ V/(kA/m}^2) = 0.360\text{ V}$:

$$\Delta U_{\text{mem}} = 0.360\text{ V}$$
$$\Delta U_{\text{sol}} = 6.00\text{ kA/m}^2 \times 0.025\text{ V/(kA/m}^2) = 0.150\text{ V}$$

Sum all components:

$$U_{\text{cell}} = E_{\text{rev}} + \eta_a + |\eta_c| + \Delta U_{\text{mem}} + \Delta U_{\text{sol}} + \Delta U_{\text{struct}}$$
$$U_{\text{cell}} = 2.160 + 0.065 + 0.095 + 0.360 + 0.150 + 0.040 = 2.870\text{ V}$$

The operating cell voltage is $2.870\text{ V}$.

Step 2: Energy Efficiency Distribution

Thermodynamic reversible fraction:

$$\% \text{ Reversible} = \frac{E_{\text{rev}}}{U_{\text{cell}}} \times 100\% = \frac{2.160}{2.870} \times 100\% = 75.26\%$$

Irreversible dissipation fraction (overpotentials + ohmic heat):

$$\% \text{ Dissipation} = 100\% - 75.26\% = 24.74\%$$

Step 3: Specific Electrical Energy Consumption

Specific electrical energy consumption ($w_{\text{spec}}$) per metric ton ($1,000\text{ kg} = 25,000\text{ mol}$) of $\text{NaOH}$:

$$w_{\text{spec}} = \frac{U_{\text{cell}} \cdot F}{M_{\text{NaOH}} \cdot \eta_{\text{NaOH}} \cdot 3,600\text{ s/h}}$$

where $F = 96,485\text{ A}\cdot\text{s/mol}$, $M_{\text{NaOH}} = 0.04000\text{ kg/mol}$.

$$w_{\text{spec}} = \frac{2.870\text{ V} \times 96,485\text{ C/mol}}{0.04000\text{ kg/mol} \times 0.965 \times 3,600\text{ s/h}} = \frac{276,911.95}{138.96} = 1,992.7\text{ kWh/kg} \text{ ??? no, convert kg to tons:}$$

Recomputing:

$$w_{\text{spec}} = \frac{2.870\text{ V} \times 96,485\text{ A}\cdot\text{s/mol} \times 25,000\text{ mol/t}}{3.6 \times 10^6\text{ J/kWh} \times 0.965} = \frac{6.9228 \times 10^9\text{ J/t}}{3.474 \times 10^6\text{ J/kWh}} = 1,992.7\text{ kWh/metric ton NaOH}$$

Specific energy consumption is $1,993\text{ kWh/t NaOH}$, showcasing the high energy efficiency of modern membrane technology.

Medium Example 8.3: Chelating Resin Ion Exchange Breakthrough & Brine Hardness

A secondary brine purification ion-exchange column contains $V_{\text{resin}} = 5.00\text{ m}^3$ of macroporous aminomethylphosphonic acid chelating resin.

  • Total operating volumetric capacity of the resin for divalent calcium ions ($\text{Ca}^{2+}$, $40.08\text{ g/mol}$) is $q_{\text{cap}} = 1.20\text{ eq/L}$ ($0.60\text{ mol Ca}^{2+}\text{/L resin}$).
  • Primary-treated feed brine flows at $\dot{V}_{\text{brine}} = 80.0\text{ m}^3\text{/h}$ and contains $3.50\text{ mg/L of Ca}^{2+}$.
  • Breakthrough occurs when $85.0\%$ of the resin column's theoretical capacity is exhausted.
  1. Calculate the total moles of $\text{Ca}^{2+}$ that can be captured prior to breakthrough.
  2. Determine the operational cycle run time of the column between regenerations in hours and days.
  3. If effluent brine during the active cycle contains $8.0\text{ ppb of Ca}^{2+}$ ($0.008\text{ mg/L}$), calculate the percentage removal efficiency.

Step 1: Usable $\text{Ca}^{2+}$ Capacity

Resin volume:

$$V_{\text{resin}} = 5.00\text{ m}^3 = 5,000\text{ L}$$

Total theoretical calcium capacity:

$$n_{\text{theo}} = 5,000\text{ L} \times 0.60\text{ mol/L} = 3,000\text{ mol }\text{Ca}^{2+}$$

Usable capacity at $85.0\%$ breakthrough threshold:

$$n_{\text{usable}} = 0.850 \times 3,000\text{ mol} = 2,550\text{ mol }\text{Ca}^{2+}$$

Step 2: Cycle Run Time Calculation

Mass of $\text{Ca}^{2+}$ entering per hour:

$$\dot{m}_{\text{Ca, in}} = 80.0\text{ m}^3\text{/h} \times 3.50\text{ g/m}^3 = 280.0\text{ g/h}$$

Moles of $\text{Ca}^{2+}$ entering per hour:

$$\dot{n}_{\text{Ca, in}} = \frac{280.0\text{ g/h}}{40.08\text{ g/mol}} = 6.986\text{ mol/h}$$

Cycle run time until breakthrough:

$$t_{\text{cycle}} = \frac{n_{\text{usable}}}{\dot{n}_{\text{Ca, in}}} = \frac{2,550\text{ mol}}{6.986\text{ mol/h}} = 365.02\text{ hours} \approx 15.21\text{ days}$$

The resin operates for $365\text{ hours}$ ($15.2\text{ days}$) before requiring acidic regeneration.

Step 3: Hardness Removal Efficiency

$$\% \text{ Removal} = \frac{C_{\text{in}} - C_{\text{out}}}{C_{\text{in}}} \times 100\% = \frac{3.50 - 0.008}{3.50} \times 100\% = \frac{3.492}{3.50} \times 100\% = 99.77\%$$

The chelating resin achieves $99.77\%$ calcium removal, lowering hardness from $3.5\text{ ppm}$ to an ultra-pure $8\text{ ppb}$.

Hard Example 8.4: Triple-Effect Caustic Soda Evaporator Material & Steam Balance

A chlor-alkali plant concentrates $\dot{m}_{\text{feed}} = 30.0\text{ metric tons/h}$ of cell liquor containing $33.0\text{ wt}\%\text{ NaOH}$ to merchant product containing $50.0\text{ wt}\%\text{ NaOH}$ in a triple-effect falling film evaporator.

  • Live motive steam ($4.0\text{ bar}$, enthalpy of vaporization $\lambda_{\text{steam}} = 2,133\text{ kJ/kg}$) is supplied to the first effect at a rate of $4,100\text{ kg/h}$.

Assume negligible solids entrainment.

  1. Calculate the production rate of $50.0\text{ wt}\%\text{ NaOH}$ solution in metric tons per hour.
  2. Determine the total hourly water evaporation rate in metric tons per hour.
  3. Calculate the overall Steam Economy of the evaporator system.

Step 1: Product Rate Calculation

NaOH mass entering in feed:

$$\dot{m}_{\text{NaOH}} = 0.330 \times 30.0\text{ metric tons/h} = 9.90\text{ metric tons/h}$$

Since all NaOH leaves in the $50.0\text{ wt}\%$ product:

$$\dot{m}_{\text{product}} = \frac{9.90\text{ metric tons/h}}{0.500} = 19.80\text{ metric tons/h}$$

The plant produces $19.80\text{ metric tons/h}$ of $50\%\text{ NaOH}$ solution.

Step 2: Water Evaporation Rate

Water entering in feed:

$$W_{\text{in}} = 30.0 - 9.90 = 20.10\text{ metric tons/h}$$

Water leaving in product:

$$W_{\text{out}} = 19.80 - 9.90 = 9.90\text{ metric tons/h}$$

Total water evaporated as vapor:

$$V_{\text{total}} = W_{\text{in}} - W_{\text{out}} = 20.10 - 9.90 = 10.20\text{ metric tons/h} = 10,200\text{ kg/h}$$

Step 3: Steam Economy

Motive steam supplied:

$$\dot{m}_{\text{steam}} = 4,100\text{ kg/h}$$

Steam economy:

$$\text{Steam Economy} = \frac{\text{Mass of water evaporated}}{\text{Mass of motive steam supplied}} = \frac{10,200\text{ kg/h}}{4,100\text{ kg/h}} = 2.488 \approx 2.49$$

The triple-effect evaporator achieves a steam economy of $2.49\text{ kg vapor / kg steam}$.

Medium Example 8.5: Chlorine Liquefaction Refrigeration Cycle & Sniff Gas Purge

A chlorine liquefaction unit receives $\dot{m}_{\text{gas}} = 5,000\text{ kg/h}$ of bone-dry chlorine gas at $8.0\text{ bar}$ absolute and $30^\circ\text{C}$.

  • Gas composition: $97.0\text{ mol}\%\text{ Cl}_2$ and $3.0\text{ mol}\%$ non-condensable inerts ($\text{O}_2, \text{N}_2, \text{H}_2$, average $M_{\text{inerts}} = 30.0\text{ g/mol}$).
  • The condenser cools the stream to $-15^\circ\text{C}$ ($258.15\text{ K}$) at constant total pressure $P = 8.0\text{ bar}$.
  • At $-15^\circ\text{C}$, the saturation vapor pressure of pure chlorine is $p_{\text{Cl}_2}^* = 1.35\text{ bar}$.
  • Latent heat of condensation of chlorine at $-15^\circ\text{C}$ is $\Delta H_{\text{cond}} = 275\text{ kJ/kg}$.
  • Specific heat of gaseous chlorine is $c_p = 0.49\text{ kJ/(kg}\cdot\text{K)}$.
  1. Calculate the molar flow rates of entering $\text{Cl}_2$ and non-condensable inerts.
  2. Determine the molar composition of the tail gas ("sniff gas") exiting the condenser and the moles of $\text{Cl}_2$ remaining in the sniff gas.
  3. Calculate the percentage of chlorine liquefied.
  4. Calculate the refrigeration cooling duty required ($\text{kW}$).

Step 1: Input Molar Flow Rates

Average molar mass of feed gas:

$$\bar{M} = 0.970(70.90) + 0.030(30.00) = 68.773 + 0.900 = 69.673\text{ g/mol}$$

Total molar feed rate:

$$\dot{n}_{\text{total}} = \frac{5,000\text{ kg/h}}{69.673\text{ kg/kmol}} = 71.764\text{ kmol/h}$$
  • Entering $\text{Cl}_2$: $\dot{n}_{\text{Cl}_2, \text{in}} = 0.970 \times 71.764 = 69.611\text{ kmol/h} = 4,935.4\text{ kg/h}$
  • Entering inerts: $\dot{n}_{\text{inerts}} = 0.030 \times 71.764 = 2.153\text{ kmol/h}$

Step 2: Sniff Gas Equilibrium & Chlorine Loss

In the sniff gas at $-15^\circ\text{C}$ and $P = 8.0\text{ bar}$:

$$y_{\text{Cl}_2} = \frac{p_{\text{Cl}_2}^*}{P} = \frac{1.35\text{ bar}}{8.00\text{ bar}} = 0.16875 \quad (16.88\%)$$

The inert fraction is $y_{\text{inerts}} = 1 - 0.16875 = 0.83125$. Since inerts do not condense, all $2.153\text{ kmol/h}$ of inerts leave in the sniff gas:

$$\dot{n}_{\text{sniff, total}} = \frac{\dot{n}_{\text{inerts}}}{y_{\text{inerts}}} = \frac{2.153\text{ kmol/h}}{0.83125} = 2.590\text{ kmol/h}$$

Chlorine lost in sniff gas:

$$\dot{n}_{\text{Cl}_2, \text{sniff}} = y_{\text{Cl}_2} \times \dot{n}_{\text{sniff, total}} = 0.16875 \times 2.590 = 0.437\text{ kmol/h}$$

Mass of $\text{Cl}_2$ lost:

$$\dot{m}_{\text{Cl}_2, \text{sniff}} = 0.437\text{ kmol/h} \times 70.90\text{ kg/kmol} = 30.98\text{ kg/h}$$

Step 3: Chlorine Liquefaction Yield

Liquefied chlorine:

$$\dot{m}_{\text{Cl}_2, \text{liquid}} = 4,935.4 - 31.0 = 4,904.4\text{ kg/h}$$

Liquefaction recovery:

$$\% \text{ Recovery} = \frac{4,904.4\text{ kg/h}}{4,935.4\text{ kg/h}} \times 100\% = 99.37\%$$

Step 4: Refrigeration Cooling Duty

  1. Sensible cooling of gas from $30^\circ\text{C}$ to $-15^\circ\text{C}$ ($\Delta T = 45\text{ K}$):
$$\dot{Q}_{\text{sens}} = 5,000\text{ kg/h} \times 0.49\text{ kJ/(kg}\cdot\text{K)} \times 45\text{ K} = 110,250\text{ kJ/h}$$
  1. Latent heat of condensation:
$$\dot{Q}_{\text{latent}} = 4,904.4\text{ kg/h} \times 275\text{ kJ/kg} = 1,348,710\text{ kJ/h}$$

Total thermal duty:

$$\dot{Q}_{\text{total}} = 110,250 + 1,348,710 = 1,458,960\text{ kJ/h}$$

In thermal kilowatts ($\text{kW}$):

$$\dot{Q}_{\text{refrig}} = \frac{1,458,960\text{ kJ/h}}{3,600\text{ s/h}} = 405.27\text{ kW}$$

The refrigeration unit delivers $405.3\text{ kW}$ of cooling, liquefying $99.37\%$ of the chlorine.

Easy Example 8.6: Synthesis of 33 wt% Hydrochloric Acid from H2 and Cl2

An $\text{HCl}$ synthesis unit reacts pure chlorine and hydrogen:

$$\text{H}_2(g) + \text{Cl}_2(g) \longrightarrow 2\text{HCl}(g)$$
  • Chlorine feed rate is $\dot{m}_{\text{Cl}_2} = 1,418\text{ kg/h}$ ($20.0\text{ kmol/h}$, $M = 70.90\text{ g/mol}$).
  • Hydrogen is supplied at a $6.0\%$ stoichiometric excess.
  • The generated anhydrous $\text{HCl}$ gas ($36.46\text{ g/mol}$) is completely absorbed into demineralized water in an isothermal falling-film graphite absorber to produce commercial $33.0\text{ wt}\%\text{ aqueous hydrochloric acid}$.
  1. Calculate the required mass feed rate of hydrogen gas in $\text{kg/h}$ ($M = 2.016\text{ g/mol}$).
  2. Determine the production rate of anhydrous $\text{HCl}$ gas in $\text{kg/h}$.
  3. Calculate the required demineralized water flow rate and the total production rate of $33.0\text{ wt}\%\text{ HCl}$ in metric tons per hour.

Step 1: Hydrogen Feed Rate

Moles of $\text{Cl}_2$ fed per hour:

$$\dot{n}_{\text{Cl}_2} = \frac{1,418\text{ kg/h}}{70.90\text{ kg/kmol}} = 20.00\text{ kmol/h}$$

With $6.0\%$ stoichiometric excess:

$$\dot{n}_{\text{H}_2} = 1.06 \times 20.00\text{ kmol/h} = 21.20\text{ kmol/h}$$

Mass feed rate of $\text{H}_2$:

$$\dot{m}_{\text{H}_2} = 21.20\text{ kmol/h} \times 2.016\text{ kg/kmol} = 42.74\text{ kg/h}$$

Step 2: Anhydrous $\text{HCl}$ Gas Production

From stoichiometry, $1\text{ mol Cl}_2 \to 2\text{ mol HCl}$:

$$\dot{n}_{\text{HCl}} = 2 \times 20.00 = 40.00\text{ kmol/h}$$

Mass of anhydrous $\text{HCl}$:

$$\dot{m}_{\text{HCl}} = 40.00\text{ kmol/h} \times 36.46\text{ kg/kmol} = 1,458.4\text{ kg/h}$$

Step 3: Demineralized Water Flow & Product Acid Rate

Target concentration is $33.0\text{ wt}\%\text{ HCl}$:

$$\dot{m}_{\text{acid 33\%}} = \frac{\dot{m}_{\text{HCl}}}{0.330} = \frac{1,458.4\text{ kg/h}}{0.330} = 4,419.4\text{ kg/h} \approx 4.419\text{ metric tons/h}$$

Demineralized water absorption rate:

$$\dot{m}_{\text{water}} = \dot{m}_{\text{acid}} - \dot{m}_{\text{HCl}} = 4,419.4 - 1,458.4 = 2,961.0\text{ kg/h} \approx 2.961\text{ metric tons/h}$$

The plant feeds $42.7\text{ kg/h}$ H2 and $2.96\text{ t/h}$ water to produce $4.42\text{ metric tons/h}$ of $33\%\text{ HCl}$ acid.

Medium Example 8.7: DCDA Contact Process Overall SO2 Conversion & Emission Compliance

A sulfuric acid plant produces $1,000\text{ metric tons/day}$ ($41.67\text{ t/h}$) of $100\%\text{ H}_2\text{SO}_4$ equivalent ($98.08\text{ g/mol}$) utilizing a $3+1$ bed Double Contact Double Absorption (DCDA) layout.

  • The sulfur burner produces process gas containing $11.0\text{ vol}\%\text{ SO}_2$ and $10.0\text{ vol}\%\text{ O}_2$.
  • In the primary contact loop (Beds 1, 2, 3), fractional conversion of $\text{SO}_2$ to $\text{SO}_3$ reaches $\alpha_1 = 94.0\%$.
  • The intermediate absorption tower (IPAT) absorbs $99.8\%$ of the generated $\text{SO}_3$.
  • In the secondary contact loop (Bed 4), the remaining unreacted $\text{SO}_2$ achieves a fractional conversion of $\alpha_2 = 98.0\%$.
  1. Calculate the overall cumulative $\text{SO}_2 \to \text{SO}_3$ conversion efficiency ($\alpha_{\text{total}}$) of the DCDA plant.
  2. Determine the mass of unreacted $\text{SO}_2$ ($64.06\text{ g/mol}$) discharged to the stack per hour in $\text{kg/h}$.
  3. Calculate the specific $\text{SO}_2$ emission per metric ton of $100\%\text{ H}_2\text{SO}_4$ produced and verify if it meets the World Bank/EPA standard ($\le 2.0\text{ kg SO}_2\text{/t acid}$).

Step 1: Overall Cumulative Conversion ($\alpha_{\text{total}}$)

Let initial moles of $\text{SO}_2$ entering Bed 1 be $n_0 = 1.000$.

  • After primary loop (Beds 1-3):
  • Converted to $\text{SO}_3$: $\alpha_1 = 0.940$
  • Remaining unreacted $\text{SO}_2$: $1 - \alpha_1 = 0.060$
  • In Bed 4, this remaining $0.060$ undergoes conversion $\alpha_2 = 0.980$:
  • Additional $\text{SO}_3$ formed: $0.060 \times 0.980 = 0.0588$
  • Final unconverted $\text{SO}_2$: $0.060 \times (1 - 0.980) = 0.060 \times 0.020 = 0.0012$

Total $\text{SO}_2$ converted across the plant:

$$\alpha_{\text{total}} = 1.000 - 0.0012 = 0.9988 \quad (99.88\%)$$

The DCDA plant achieves an overall conversion efficiency of $99.88\%$.

Step 2: Unreacted $\text{SO}_2$ Discharged to Stack

Hourly production of $100\%\text{ H}_2\text{SO}_4$:

$$\dot{m}_{\text{acid}} = 41.667\text{ metric tons/h} = 41,667\text{ kg/h}$$

Moles of sulfuric acid produced per hour:

$$\dot{n}_{\text{acid}} = \frac{41,667\text{ kg/h}}{98.08\text{ kg/kmol}} = 424.83\text{ kmol/h}$$

Since $1\text{ mole of converted SO}_2$ yields $1\text{ mole of H}_2\text{SO}_4$:

$$\dot{n}_{\text{SO}_2, \text{converted}} = 424.83\text{ kmol/h}$$

Initial $\text{SO}_2$ burned:

$$\dot{n}_{\text{SO}_2, \text{total fed}} = \frac{424.83\text{ kmol/h}}{0.9988} = 425.34\text{ kmol/h}$$

Unreacted $\text{SO}_2$ exiting to stack:

$$\dot{n}_{\text{SO}_2, \text{stack}} = 425.34 - 424.83 = 0.510\text{ kmol/h}$$

Mass of stack $\text{SO}_2$:

$$\dot{m}_{\text{SO}_2, \text{stack}} = 0.510\text{ kmol/h} \times 64.06\text{ kg/kmol} = 32.67\text{ kg/h}$$

Step 3: Specific Emission Compliance

Specific emission factor:

$$E_{\text{spec}} = \frac{32.67\text{ kg SO}_2\text{/h}}{41.667\text{ t acid/h}} = 0.784\text{ kg SO}_2\text{/metric ton acid}$$

Since $0.784\text{ kg/t} < 2.0\text{ kg/t}$, the plant easily meets the regulatory limit with a $61\%$ safety margin.

Easy Example 8.8: Oxygen-Depolarized Cathode (ODC) Voltage & Energy Conservation

A chlor-alkali plant operates at an electrical current of $I = 100\text{ kA}$ ($100,000\text{ A}$) to produce $100\%\text{ pure NaOH}$ ($40.00\text{ g/mol}$) at $\eta = 96.0\%$ current efficiency ($F = 96,485\text{ C/mol}$).

  1. In a conventional hydrogen-evolving membrane cell, the operational cell voltage is $U_1 = 3.050\text{ V}$.
  2. In an advanced Oxygen-Depolarized Cathode (ODC) cell, pure oxygen gas is supplied to the cathode, reducing the operational cell voltage to $U_2 = 2.050\text{ V}$.

Industrial electricity costs $\$0.080\text{ per kWh}$.

  1. Calculate the hourly direct-current electrical power consumption of a single cell under:
  • Conventional cell ($U_1 = 3.050\text{ V}$).
  • ODC cell ($U_2 = 2.050\text{ V}$).
  1. Determine the specific direct-current energy consumption per metric ton of pure $\text{NaOH}$ for both technologies ($\text{kWh/t NaOH}$).
  2. Calculate the percentage electrical energy savings and the annual electricity cost savings per cell (operating $8,400\text{ hours/year}$).

Step 1: Hourly Electrical Power per Cell

Power is $P = U \times I$:

  • Conventional Cell:
$$P_1 = 3.050\text{ V} \times 100,000\text{ A} = 305,000\text{ W} = 305.0\text{ kW}$$
  • ODC Cell:
$$P_2 = 2.050\text{ V} \times 100,000\text{ A} = 205,000\text{ W} = 205.0\text{ kW}$$

Step 2: Specific Energy Consumption per Metric Ton NaOH

Hourly NaOH production per cell ($M = 40.00\text{ g/mol}$):

$$\dot{n}_{\text{NaOH}} = \frac{100,000\text{ C/s} \times 3,600\text{ s/h} \times 0.960}{96,485\text{ C/mol}} = 3,582.0\text{ mol/h}$$
$$\dot{m}_{\text{NaOH}} = 3,582.0\text{ mol/h} \times 0.04000\text{ kg/mol} = 143.28\text{ kg/h} = 0.14328\text{ metric tons/h}$$

Specific energy consumption ($w_{\text{spec}} = P / \dot{m}$):

  • Conventional:
$$w_{\text{spec}, 1} = \frac{305.0\text{ kW}}{0.14328\text{ t/h}} = 2,128.7\text{ kWh/metric ton NaOH}$$
  • ODC Cell:
$$w_{\text{spec}, 2} = \frac{205.0\text{ kW}}{0.14328\text{ t/h}} = 1,430.8\text{ kWh/metric ton NaOH}$$

Step 3: Savings Analysis

Percentage energy savings:

$$\% \text{ Savings} = \frac{3.050 - 2.050}{3.050} \times 100\% = \frac{1.000}{3.050} \times 100\% = 32.79\%$$

Annual electrical energy saved per cell ($8,400\text{ operating hours}$):

$$\Delta E = (305.0 - 205.0\text{ kW}) \times 8,400\text{ h} = 100.0\text{ kW} \times 8,400\text{ h} = 840,000\text{ kWh/year}$$

Annual financial savings per cell:

$$\text{Cost Savings} = 840,000\text{ kWh} \times \$0.080\text{/kWh} = \$67,200\text{/cell/year}$$

ODC technology slashes power consumption by $32.8\%$, saving $\$67,200\text{ per cell annually}$.

Hard Example 8.9: Bipolar Membrane Electrodialysis (BMED) Water Splitting

A chemical zero-liquid-discharge facility treats industrial waste sodium sulfate ($\text{Na}_2\text{SO}_4$, $142.04\text{ g/mol}$) using Bipolar Membrane Electrodialysis (BMED) to regenerate sodium hydroxide ($\text{NaOH}$, $40.00\text{ g/mol}$) and sulfuric acid ($\text{H}_2\text{SO}_4$, $98.08\text{ g/mol}$).

  • The BMED stack comprises $N = 200\text{ repeating cell triplets}$ (Bipolar Membrane - Anion Exchange Membrane - Cation Exchange Membrane) operating in electrical series at $I = 500\text{ A}$.
  • At the bipolar membrane junction, water dissociates into $\text{H}^+$ and $\text{OH}^-$ ions:
$$\text{H}_2\text{O} \overset{\text{Electrical Field}}{\longrightarrow} \text{H}^+ + \text{OH}^-$$
  • The current efficiency for $\text{NaOH}$ is $\eta = 88.0\%$ ($F = 96,485\text{ C/mol}$).
  • The overall cell triplet operating voltage is $U = 1.80\text{ V}$.
  1. Calculate the daily production rate of pure $100\%\text{ NaOH}$ in metric tons per day.
  2. Determine the daily production rate of pure $\text{H}_2\text{SO}_4$ generated in metric tons per day.
  3. Calculate the specific direct-current electrical energy consumption per metric ton of $\text{NaOH}$ produced ($\text{kWh/t NaOH}$).

Step 1: Daily $\text{NaOH}$ Production

Total charge passed across $N = 200$ cells in series per day ($t = 86,400\text{ s}$):

$$Q_{\text{day}} = 200 \times 500\text{ A} \times 86,400\text{ s} = 8.64 \times 10^9\text{ C}$$

Moles of electrons passed:

$$n_e = \frac{8.64 \times 10^9\text{ C}}{96,485\text{ C/mol}} = 89,547.6\text{ mol } e^- = 89.548\text{ kmol } e^-$$

At $\eta = 88.0\%$ current efficiency, moles of $\text{NaOH}$ generated:

$$n_{\text{NaOH}} = 0.880 \times 89.548\text{ kmol} = 78.802\text{ kmol/day}$$

Mass of pure $\text{NaOH}$:

$$m_{\text{NaOH}} = 78.802\text{ kmol} \times 40.00\text{ kg/kmol} = 3,152.1\text{ kg/day} \approx 3.152\text{ metric tons/day}$$

Step 2: Daily $\text{H}_2\text{SO}_4$ Production

Each mole of $\text{H}_2\text{SO}_4$ requires $2\text{ moles of H}^+$:

$$n_{\text{H}_2\text{SO}_4} = \frac{n_{\text{NaOH}}}{2} = \frac{78.802}{2} = 39.401\text{ kmol/day}$$

Mass of pure $\text{H}_2\text{SO}_4$:

$$m_{\text{H}_2\text{SO}_4} = 39.401\text{ kmol} \times 98.08\text{ kg/kmol} = 3,864.5\text{ kg/day} \approx 3.865\text{ metric tons/day}$$

Step 3: Specific Electrical Energy Consumption

Total direct-current electrical power of the stack ($200\text{ triplets}$ at $1.80\text{ V}$, total voltage $= 360\text{ V}$):

$$P = 200 \times 1.80\text{ V} \times 500\text{ A} = 180,000\text{ W} = 180.0\text{ kW}$$

Daily energy consumed:

$$E_{\text{day}} = 180.0\text{ kW} \times 24\text{ h} = 4,320\text{ kWh/day}$$

Specific energy consumption per ton of $\text{NaOH}$:

$$w_{\text{spec}} = \frac{4,320\text{ kWh/day}}{3.1521\text{ t NaOH/day}} = 1,370.5\text{ kWh/metric ton NaOH}$$

BMED produces $3.15\text{ t/day}$ NaOH and $3.86\text{ t/day}$ H2SO4, consuming $1,371\text{ kWh/t NaOH}$.

Solved Honors Problems & Derivations

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