Environmental & Industrial Chemistry

Industrial Processes: The Haber, Contact, and Chlor-alkali Processes

5th Year · 6th Year (Leaving Cert)

✓By the end of this lesson students will be able to outline the principles, raw materials, conditions, and products of the Haber, Contact, and Chlor-alkali processes.
✓By the end of this lesson students will be able to explain the application of Le Chatelier's Principle to optimise yields in the Haber and Contact processes.
✓By the end of this lesson students will be able to detail the reactions occurring at the electrodes in the Chlor-alkali process and compare different cell types.
✓By the end of this lesson students will be able to describe the environmental considerations associated with each industrial process.

Key concepts

The Haber Process

The Haber process is the industrial synthesis of ammonia (NH₃) from nitrogen (N₂) and hydrogen (H₂). It is a reversible, exothermic reaction. To achieve a high yield and a reasonable reaction rate, specific conditions are employed based on Le Chatelier's Principle. Nitrogen is obtained from the air, and hydrogen is typically produced from natural gas (methane) by steam reforming. Ammonia is a crucial raw material for fertilisers and the production of nitric acid.

N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol⁻¹

The Contact Process

The Contact process is the industrial method for manufacturing sulfuric acid (H₂SO₄), one of the most important industrial chemicals. The key steps involve the production of sulfur dioxide (SO₂), the catalytic oxidation of SO₂ to sulfur trioxide (SO₃), and the absorption of SO₃. The catalytic step is a reversible, exothermic reaction, and conditions are chosen to maximise yield and rate. Sulfuric acid is used extensively in fertiliser production, detergents, and many other chemical industries.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = -197 kJ mol⁻¹

The Chlor-alkali Process

The Chlor-alkali process is the electrolysis of concentrated aqueous sodium chloride (brine) to produce chlorine gas (Cl₂), hydrogen gas (H₂), and sodium hydroxide (NaOH). These three products are vital industrial chemicals. Different types of electrolytic cells are used, primarily diaphragm cells and membrane cells. Membrane cells are now preferred due to their higher energy efficiency and the production of purer sodium hydroxide, as well as being more environmentally friendly by avoiding asbestos or mercury.

2NaCl(aq) + 2H₂O(l) → 2NaOH(aq) + Cl₂(g) + H₂(g)

Key facts to remember

1The Haber process uses N₂ from air and H₂ from natural gas to produce ammonia (NH₃).
2Optimal conditions for the Haber process are 400-450 °C, 200-250 atm, and a finely divided iron catalyst.
3The Contact process uses sulfur (or sulfide ores) and air to produce sulfuric acid (H₂SO₄).
4The catalytic step in the Contact process (SO₂ to SO₃) uses a vanadium(V) oxide (V₂O₅) catalyst at approximately 450 °C and 1-2 atm.
5In the Contact process, SO₃ is absorbed into concentrated H₂SO₄ to form oleum (H₂S₂O₇), preventing corrosive mist formation.
6The Chlor-alkali process is the electrolysis of concentrated aqueous NaCl (brine) to produce Cl₂, H₂, and NaOH.
7Membrane cells are the preferred modern method for the Chlor-alkali process due to their efficiency and environmental benefits.
8All three processes are energy-intensive and have significant environmental considerations, such as energy consumption and potential pollution.

Worked examples

Example 1

The Haber process for the synthesis of ammonia is represented by the equation: N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = -92 kJ mol⁻¹. Explain, using Le Chatelier's Principle, the effect of (a) increasing the temperature and (b) increasing the pressure on the equilibrium yield of ammonia.

I(a) Effect of increasing temperature:

II1. Identify the nature of the forward reaction: The forward reaction (formation of ammonia) is exothermic, as indicated by ΔH = -92 kJ mol⁻¹.

III2. Apply Le Chatelier's Principle: If the temperature is increased, the equilibrium will shift in the endothermic direction to absorb the added heat.

IV3. Conclusion: The reverse reaction (decomposition of ammonia) is endothermic and will be favoured. This leads to a decrease in the equilibrium yield of ammonia.

V(b) Effect of increasing pressure:

VI1. Count the total moles of gaseous reactants and products: On the reactant side, there are 1 mole of N₂ + 3 moles of H₂ = 4 moles of gas. On the product side, there are 2 moles of NH₃ = 2 moles of gas.

VII2. Apply Le Chatelier's Principle: If the pressure is increased, the equilibrium will shift to the side with fewer moles of gas to reduce the pressure.

VIII3. Conclusion: The forward reaction (formation of ammonia) produces fewer moles of gas (2 moles) than the reactants (4 moles). Therefore, increasing the pressure favours the forward reaction, leading to an increase in the equilibrium yield of ammonia.

Answer

(a) Increasing the temperature decreases the equilibrium yield of ammonia. (b) Increasing the pressure increases the equilibrium yield of ammonia.

In practice, a compromise temperature (400-450 °C) is used to achieve a reasonable reaction rate, despite a lower equilibrium yield at higher temperatures.

Example 2

In the Contact process, sulfur trioxide (SO₃) is absorbed into concentrated sulfuric acid (H₂SO₄) rather than directly into water. Explain why this is the case, including relevant equations.

I1. Problem with direct absorption into water: If sulfur trioxide (SO₃) were absorbed directly into water (H₂O), the reaction SO₃(g) + H₂O(l) → H₂SO₄(aq) is highly exothermic.

II2. Consequence: This highly exothermic reaction would produce a dense, corrosive mist of fine sulfuric acid droplets that are very difficult to condense and handle safely, posing significant health and safety hazards.

III3. Industrial solution: To avoid this mist formation, SO₃ is absorbed into concentrated sulfuric acid (H₂SO₄) to form oleum (fuming sulfuric acid), which has the formula H₂S₂O₇.

IV4. Equation for absorption: SO₃(g) + H₂SO₄(l) → H₂S₂O₇(l)

V5. Subsequent step: The oleum is then safely diluted with a carefully controlled amount of water to produce sulfuric acid of the desired concentration.

VI6. Equation for dilution: H₂S₂O₇(l) + H₂O(l) → 2H₂SO₄(l)

Answer

Direct absorption of SO₃ into water is too exothermic, producing a dangerous, corrosive mist of H₂SO₄. Instead, SO₃ is absorbed into concentrated H₂SO₄ to form oleum (H₂S₂O₇), which is then safely diluted with water to yield H₂SO₄.

Example 3

In a membrane cell used for the Chlor-alkali process, identify the products formed at the anode and cathode, and write the balanced half-equations for the reactions occurring at each electrode.

I1. Identify the electrolyte and species present: The electrolyte is concentrated aqueous sodium chloride (brine), so Na⁺, Cl⁻, H₂O, and a small amount of OH⁻ (from water dissociation) are present.

II2. Anode (positive electrode): Oxidation occurs here. Possible species to be oxidised are Cl⁻ and H₂O. Due to the high concentration of Cl⁻ ions and the overpotential for oxygen formation, chloride ions are preferentially oxidised.

III3. Anode half-equation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻. The product is chlorine gas.

IV4. Cathode (negative electrode): Reduction occurs here. Possible species to be reduced are Na⁺ and H₂O. Water is more easily reduced than sodium ions.

V5. Cathode half-equation: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq). The products are hydrogen gas and hydroxide ions.

VI6. Overall products: Chlorine gas (Cl₂) is formed at the anode. Hydrogen gas (H₂) and hydroxide ions (OH⁻) are formed at the cathode. The hydroxide ions combine with the remaining sodium ions (Na⁺) to form sodium hydroxide (NaOH) in the cathode compartment.

Answer

At the anode: Chlorine gas (Cl₂) is formed. Half-equation: 2Cl⁻(aq) → Cl₂(g) + 2e⁻. At the cathode: Hydrogen gas (H₂) and hydroxide ions (OH⁻) are formed. Half-equation: 2H₂O(l) + 2e⁻ → H₂(g) + 2OH⁻(aq).

The membrane in a membrane cell allows only Na⁺ ions to pass through, ensuring that the NaOH produced is pure and concentrated, without contamination by Cl⁻ ions.

Common mistakes

✗Confusing the optimal temperature and pressure conditions for the Haber and Contact processes.
✗Incorrectly applying Le Chatelier's Principle, especially regarding the effect of a catalyst (which affects rate but not equilibrium position).
✗Stating that SO₃ is absorbed directly into water in the Contact process, rather than into concentrated H₂SO₄.
✗Incorrectly identifying the products or writing the half-equations for the anode and cathode in the Chlor-alkali process.
✗Overlooking the environmental impact and sustainability aspects of these industrial processes in exam answers.

Exam tips

★Memorise the key raw materials, catalysts, and operating conditions (temperature and pressure) for each process.
★Be able to clearly explain and apply Le Chatelier's Principle to the Haber and Contact processes, justifying the chosen conditions.
★Practise writing balanced chemical equations for all main reactions, including half-equations for the Chlor-alkali process.
★Understand the reasons behind specific steps (e.g., why SO₃ is absorbed into H₂SO₄) and the advantages/disadvantages of different cell types in the Chlor-alkali process.
★Be prepared to discuss the economic and environmental implications of each process, including energy usage and pollution control.

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