HomeChemistry › Q38
Chemistryd-block Elements
Match List I (Catalyst) with List II (Process/Reaction):

List I (Catalyst)

A. V₂O₅
B. Fe
C. PdCl₂
D. Ni

List II (Process)

I. Haber process
II. Hydrogenation of oils
III. Contact process
IV. Wacker process
Options
1
A–III, B–I, C–II, D–IV
2
A–III, B–IV, C–I, D–II
3
A–I, B–III, C–IV, D–II
4
A–III, B–I, C–IV, D–II
Correct Answer
A–III, B–I, C–IV, D–II
Solution — Each Match Explained
A

V₂O₅ → III. Contact Process (H₂SO₄ manufacture)

Vanadium pentoxide oxidises SO₂ to SO₃, cycling between V⁵⁺ (V₂O₅) and V⁴⁺ (V₂O₄). Conditions: 450°C, 1–2 atm pressure.

2SO₂ + O₂ → 2SO₃   (V₂O₅ catalyst)

B

Fe → I. Haber Process (NH₃ manufacture)

Finely divided iron with Mo as promoter. N₂ and H₂ adsorb on Fe surface; N≡N bond weakens, enabling reaction.

N₂ + 3H₂ ⇌ 2NH₃   (Fe catalyst, 400–500°C, 200 atm)

C

PdCl₂ → IV. Wacker Process (acetaldehyde from ethylene)

PdCl₂ oxidises ethylene to acetaldehyde; Pd²⁺ is reduced to Pd⁰, then regenerated by CuCl₂/O₂.

CH₂=CH₂ + ½O₂ → CH₃CHO   (PdCl₂/CuCl₂)

D

Ni → II. Hydrogenation of oils (vanaspati ghee)

Raney nickel (finely divided, high surface area) adsorbs H₂ and transfers it to C=C double bonds in unsaturated oils.

Vegetable oil + H₂ → Saturated fat   (Ni, ~180°C)

Theory: Transition Metals as Catalysts
1. Why Transition Metals Make Excellent Catalysts

Transition metals are exceptional catalysts because of their unique electronic structure — partially filled d-orbitals. This gives them two key catalytic abilities: variable oxidation states (they can gain and lose electrons reversibly, enabling redox catalysis) and the ability to form coordinate bonds with reactant molecules (enabling surface adsorption and intermediate formation).

In heterogeneous catalysis, the metal surface provides active sites where reactant molecules adsorb, have their bonds weakened, react with each other, and then desorb as products. The catalyst is regenerated in the process. The key is that the metal-reactant bond must be strong enough to hold reactants but weak enough to release products — this is the Sabatier principle, and transition metals near the middle of the d-block (Fe, Ni, Pt, Pd) satisfy it best.

2. Contact Process — Complete Details

Sulphuric acid (H₂SO₄) is the most produced industrial chemical in the world, with about 200 million tonnes made annually. It is used to make fertilisers, detergents, dyes, drugs, and hundreds of other chemicals. The Contact process involves three key steps:

📌 Step 1: Burn sulphur → S + O₂ → SO₂

📌 Step 2: Oxidise SO₂ → 2SO₂ + O₂ → 2SO₃ (V₂O₅, 450°C)

📌 Step 3: Dissolve in H₂SO₄ → SO₃ + H₂SO₄ → H₂S₂O₇ (oleum)

📌 Step 4: Dilute oleum → H₂S₂O₇ + H₂O → 2H₂SO₄

📌 V₂O₅ cycle: V⁵⁺ + SO₂ → V⁴⁺ + SO₃; V⁴⁺ + ½O₂ → V⁵⁺

📌 Direct dissolution of SO₃ in water is not done (fog formation)

3. Haber Process — Complete Details

The Haber process for ammonia synthesis is one of the most important chemical reactions in history — it enabled nitrogen fixation at industrial scale, making artificial fertilisers possible and supporting the food needs of billions of people. Fritz Haber won the Nobel Prize in 1918 for developing it.

📌 Reaction: N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ/mol (exothermic)

📌 Temperature: 400–500°C (compromise — faster rate vs higher yield)

📌 Pressure: 150–200 atm (high P favours NH₃ — fewer moles of gas)

📌 Catalyst: Fe (finely divided), Promoters: Mo, Al₂O₃, K₂O

📌 Yield: only ~15% per pass; unreacted N₂/H₂ recycled

📌 N₂ source: atmosphere; H₂ source: steam reforming of methane

4. Wacker Process — Mechanism

The Wacker process was developed in the 1950s and revolutionised production of acetaldehyde. The full catalytic cycle involves three coupled redox reactions:

📌 Main: C₂H₄ + PdCl₂ + H₂O → CH₃CHO + Pd⁰ + 2HCl

📌 Regenerate Pd: Pd⁰ + 2CuCl₂ → PdCl₂ + 2CuCl

📌 Regenerate Cu: 2CuCl + ½O₂ + 2HCl → 2CuCl₂ + H₂O

📌 Net: CH₂=CH₂ + ½O₂ → CH₃CHO (ethylene → acetaldehyde)

📌 Pd and Cu are not consumed — they are true catalysts

5. Hydrogenation — Mechanism on Ni Surface

The hydrogenation mechanism on nickel surface follows the Langmuir-Hinshelwood model: (1) H₂ adsorbs on Ni surface and dissociates into two H atoms strongly bound to surface Ni atoms. (2) The alkene approaches and adsorbs on the surface. (3) One H atom is transferred from the surface to one carbon of the double bond (half-hydrogenated intermediate). (4) The second H atom transfers to the adjacent carbon. (5) The fully saturated alkane desorbs from the surface. This is why partial hydrogenation is possible — the half-hydrogenated intermediate can sometimes desorb before the second H is added, creating a mix of saturated and unsaturated products.

6. Other Important Industrial Catalysts — NEET Table

📌 Pt/Rh gauze → Ostwald process (NH₃ → NO → HNO₃)

📌 MnO₂ → KClO₃ decomposition → O₂ release

📌 Fe₂O₃ + Cr₂O₃ → water-gas shift reaction

📌 Pt/Pd → catalytic converters in cars (CO, NOₓ removal)

📌 AlCl₃ → Friedel-Crafts reactions (Lewis acid)

📌 TiCl₄/AlEt₃ → Ziegler-Natta (polyethylene, polypropylene)

📌 H₃PO₄ → industrial hydration of ethylene to ethanol

📌 ZnO/Cr₂O₃ → methanol synthesis (CO + 2H₂ → CH₃OH)

7. Promoters, Poisons, and Inhibitors

Promoters enhance catalyst activity without themselves being catalysts. Mo in the Haber process prevents Fe particles from sintering at high temperatures, maintaining high surface area. K₂O increases electron density on Fe surface, improving N₂ adsorption. Al₂O₃ acts as a structural promoter, keeping Fe particles separated.

Catalyst poisons reduce or destroy activity. Arsenic, phosphorus, and sulphur compounds poison the Fe catalyst in Haber process. Carbon monoxide poisons Pt in fuel cells and automotive catalysts. Lead compounds in petrol poisoned Pt/Pd catalytic converters (reason for switch to unleaded petrol). Enzyme inhibitors in biology work similarly — nerve agents (sarin, VX) irreversibly inhibit acetylcholinesterase.

8. Homogeneous vs Heterogeneous Catalysis

In heterogeneous catalysis (Fe, V₂O₅, Ni), the solid catalyst and gaseous/liquid reactants are in different phases. The reaction occurs at the catalyst surface. Advantages: easy separation, continuous operation, high temperature use. Disadvantage: mass transfer limitations, surface poisoning.

In homogeneous catalysis (PdCl₂ in aqueous Wacker process), catalyst and reactants are in the same phase (solution). Advantages: every catalyst molecule is accessible, mechanistic study possible. Disadvantages: separation of catalyst from product is difficult and costly. Most modern research focuses on making homogeneous catalysts recyclable — by immobilising them on solid supports (heterogenising homogeneous catalysts).

Frequently Asked Questions
1. What is the role of Mo in the Haber process?
Molybdenum (Mo) is a promoter — it does not catalyse the reaction itself but increases Fe catalyst efficiency. Mo prevents sintering (merging of Fe particles at high temperature) which would reduce surface area. It also modifies the electronic properties of Fe surface, improving N₂ adsorption and dissociation. Al₂O₃ and K₂O are also used as promoters.
2. Why is the temperature 450°C specifically chosen for V₂O₅?
At temperatures below 400°C, V₂O₅ is not active enough — the reaction is too slow. Above 600°C, the equilibrium shifts toward SO₂ (Le Chatelier — exothermic reaction disfavoured at high T), reducing SO₃ yield. 450°C is a compromise giving acceptable rate AND reasonable yield. The process gives ~98% conversion at this temperature.
3. What is Raney nickel and how is it prepared?
Raney nickel is made by: (1) Making a 50:50 alloy of Ni and Al. (2) Treating with concentrated NaOH solution: 2Al + 2NaOH + 2H₂O → 2NaAlO₂ + 3H₂. (3) Al dissolves, leaving behind porous sponge-like Ni with enormous surface area (~100 m²/g). This high surface area makes it an excellent catalyst for hydrogenation.
4. Why is SO₃ absorbed in H₂SO₄ and not directly in water?
Dissolving SO₃ directly in water would produce a fine mist of H₂SO₄ droplets (acid fog/aerosol) that is very difficult to collect and would escape as pollution. Absorbing SO₃ in concentrated H₂SO₄ gives oleum (H₂S₂O₇), a liquid that is then safely diluted with water to give H₂SO₄. This avoids the fog problem and allows efficient acid collection.
5. What products does the Wacker process ultimately lead to?
Wacker process makes acetaldehyde (CH₃CHO) from ethylene. Acetaldehyde is further oxidised to: (1) Acetic acid (CH₃COOH) — used for vinegar, solvents, vinyl acetate monomer. (2) Acetic anhydride — used for aspirin, cellulose acetate (photographic film). (3) Ethyl acetate (with ethanol) — solvent for nail polish, glues. The Wacker process thus connects cheap petroleum-derived ethylene to many important chemical intermediates.
6. What is the Ostwald process and how does it relate to Haber?
The Ostwald process converts ammonia (from Haber process) into nitric acid. Step 1: 4NH₃ + 5O₂ → 4NO + 6H₂O (Pt-Rh gauze catalyst, 850°C). Step 2: 2NO + O₂ → 2NO₂. Step 3: 3NO₂ + H₂O → 2HNO₃ + NO. HNO₃ is then mixed with NH₃ to make ammonium nitrate — an important nitrogen fertiliser. The Haber and Ostwald processes together form the industrial nitrogen cycle.
7. Why do catalytic converters use Pt and Pd?
Car exhaust catalytic converters use Pt (oxidation) and Pd (oxidation) and Rh (reduction) on a ceramic honeycomb support coated with Al₂O₃. Pt and Pd oxidise CO → CO₂ and unburnt hydrocarbons → CO₂ + H₂O. Rh reduces NOₓ → N₂. These metals are used because they have optimal adsorption energies for the relevant molecules. They are expensive, which is why car catalysts are recovered and recycled.
8. What is the Sabatier principle in catalysis?
The Sabatier principle states that a good catalyst should bind reactants strongly enough to activate them but weakly enough to release products. If binding is too weak, reactants don't adsorb and react. If too strong, products can't desorb and catalyst becomes blocked (poisoned). Transition metals in the middle of the d-block (Fe, Ni, Ru, Pd, Pt) satisfy this balance best — that's why they dominate industrial catalysis.
Previous Questions
Q.
Nitriles converted to primary amines – which reagents A B C D are correct
Amines · Answer: A, C and D only
Q.
Match EM waves with sources – Microwave, X-ray, Gamma, Infrared
EM Waves · Answer: A-IV, B-I, C-II, D-III
Q.
Submarine withstands 100 atm – maximum depth below water surface
Fluids · Answer: 990 m
Q.
Metal work function 6.6 eV – which wavelength does not give photoelectric effect
Dual Nature · Answer: 200 nm
Q.
Electrostatics statements about conductor – E=0, charge on surface, field normal
Electrostatics · Answer: A, C and D only