Industrial Chemistry — Concepts, Formulas & Examples

Industrial chemistry is how chemical knowledge becomes economically useful. CBSE Class 12 covers several key industrial processes. NEET asks direct questions on the conditions and catalysts of each.

Core Concepts

Haber process (ammonia)

$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ . Catalyst — Fe with Mo promoter. Conditions — 200 atm, 450°C. Basis of nitrogen fertilizer. Consumes about 1% of global energy.

Why these conditions?

The forward reaction is exothermic ( $\Delta H = -92$ kJ/mol) and decreases volume (4 mol gas → 2 mol gas).

By Le Chatelier’s principle:

Low temperature favours forward reaction (exothermic), but the rate becomes too slow below 400°C
High pressure favours forward reaction (fewer moles of gas), hence 200 atm
Temperature of 450°C is a compromise — fast enough kinetically, decent equilibrium yield (~15%)
The catalyst (iron) makes the reaction fast enough at 450°C to be practical

K_p = \frac{p_{\text{NH}_3}^2}{p_{\text{N}_2} \cdot p_{\text{H}_2}^3}

At 450°C and 200 atm, $K_p$ is small, giving only about 15% yield per pass. Unreacted N $_2$ and H $_2$ are recycled — the overall conversion reaches 98%.

Role of the promoter: Molybdenum (Mo) acts as a promoter — it does not catalyse the reaction itself but increases the efficiency of the iron catalyst by preventing sintering (clumping) of iron particles at high temperature.

Contact process (sulphuric acid)

$\text{S} + \text{O}_2 \to \text{SO}_2 \to \text{SO}_3 \to \text{H}_2\text{SO}_4$ . Catalyst — V $_2$ O $_5$ . Conditions — 450°C, 2 atm. Sulphuric acid is the most-produced industrial chemical.

Three stages:

\text{S} + \text{O}_2 \rightarrow \text{SO}_2

Or from pyrites: $4\text{FeS}_2 + 11\text{O}_2 \rightarrow 2\text{Fe}_2\text{O}_3 + 8\text{SO}_2$

2\text{SO}_2 + \text{O}_2 \rightleftharpoons 2\text{SO}_3 \quad (\Delta H = -196\;\text{kJ/mol})

Catalyst: V $_2$ O $_5$ at 450°C, 2 atm. Yield: ~99.5%.

SO $_3$ is NOT dissolved directly in water (this produces a dangerous mist). Instead:

\text{SO}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{H}_2\text{S}_2\text{O}_7 \text{ (oleum)}

\text{H}_2\text{S}_2\text{O}_7 + \text{H}_2\text{O} \rightarrow 2\text{H}_2\text{SO}_4

Why only 2 atm? The forward reaction reduces volume (3 mol → 2 mol), so high pressure helps. But at 450°C with V $_2$ O $_5$ , the conversion is already 99.5% at 2 atm. Higher pressure is unnecessary and would increase equipment costs.

Ostwald process (nitric acid)

Starts with NH $_3$ from Haber. $\text{NH}_3 + \text{O}_2 \to \text{NO} \to \text{NO}_2 \to \text{HNO}_3$ . Catalyst — Pt/Rh gauze. Used for fertilizers and explosives.

Three stages:

4\text{NH}_3 + 5\text{O}_2 \xrightarrow{\text{Pt/Rh, 850°C}} 4\text{NO} + 6\text{H}_2\text{O}

2\text{NO} + \text{O}_2 \rightarrow 2\text{NO}_2 \text{ (cooled to ~50°C)}

3\text{NO}_2 + \text{H}_2\text{O} \rightarrow 2\text{HNO}_3 + \text{NO} \text{ (the NO is recycled)}

The Pt/Rh catalyst is used as a fine gauze to maximise surface area. At 850°C, the selectivity to NO (desired) over N $_2$ (undesired) is maximised. The catalyst is expensive but lasts for months.

Solvay process (sodium carbonate)

$\text{NaCl} + \text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O} \to \text{NaHCO}_3 + \text{NH}_4\text{Cl}$ . NaHCO $_3$ is decomposed to Na $_2$ CO $_3$ . Used in glass, detergents and paper.

Detailed steps:

\text{NaCl} + \text{NH}_3 + \text{CO}_2 + \text{H}_2\text{O} \rightarrow \text{NaHCO}_3 \downarrow + \text{NH}_4\text{Cl}

NaHCO $_3$ precipitates because it is least soluble among the four possible salts (NaCl, NaHCO $_3$ , NH $_4$ Cl, NH $_4$ HCO $_3$ ). This is filtered and heated:

2\text{NaHCO}_3 \xrightarrow{\Delta} \text{Na}_2\text{CO}_3 + \text{H}_2\text{O} + \text{CO}_2 \uparrow

The CO $_2$ is recycled. The NH $_3$ is recovered by treating NH $_4$ Cl with Ca(OH) $_2$ :

2\text{NH}_4\text{Cl} + \text{Ca(OH)}_2 \rightarrow 2\text{NH}_3 \uparrow + \text{CaCl}_2 + 2\text{H}_2\text{O}

Elegance of the Solvay process: Only NaCl and CaCO $_3$ (source of both CO $_2$ and Ca(OH) $_2$ ) are consumed. NH $_3$ and CO $_2$ are recycled. The only by-product is CaCl $_2$ — an excellent example of industrial efficiency.

The Solvay process cannot make potassium carbonate (K $_2$ CO $_3$ ) because KHCO $_3$ is too soluble to precipitate out. K $_2$ CO $_3$ is made by other methods — a favourite tricky question.

Frasch process (sulphur)

Hot water melts underground sulphur; compressed air pushes it up. Simple mining method that gives pure sulphur.

Three concentric pipes are drilled into the sulphur deposit. Superheated water (170°C, 10 atm) is pumped down the outermost pipe, melting the sulphur (mp 115°C). Compressed air is sent down the innermost pipe, creating a froth of molten sulphur and air that rises through the middle pipe. The sulphur collected is 99.5% pure — no further purification needed.

Chlor-alkali process

Electrolysis of brine gives Cl $_2$ , H $_2$ and NaOH. Three valuable products from one process. The basis of the PVC and paper industries.

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

At the anode: $2\text{Cl}^- \rightarrow \text{Cl}_2 + 2e^-$

At the cathode: $2\text{H}_2\text{O} + 2e^- \rightarrow \text{H}_2 + 2\text{OH}^-$

The Cl $_2$ and NaOH must be kept separate (they react to form bleach). Modern membrane cells use a Nafion membrane to allow Na $^+$ migration but prevent Cl $_2$ and OH $^-$ mixing.

Summary Table

Process	Product	Catalyst	Temperature	Pressure	Key fact
Haber	NH $_3$	Fe/Mo	450°C	200 atm	15% yield per pass, recycled
Contact	H $_2$ SO $_4$	V $_2$ O $_5$	450°C	2 atm	SO $_3$ dissolved in H $_2$ SO $_4$ , not water
Ostwald	HNO $_3$	Pt/Rh	850°C	~10 atm	NH $_3$ from Haber is the starting material
Solvay	Na $_2$ CO $_3$	None	—	—	Cannot make K $_2$ CO $_3$
Chlor-alkali	Cl $_2$ , NaOH, H $_2$	—	—	—	Three products from one electrolysis

Worked Examples

Iron is cheap, effective, and stable at the high temperature and pressure used. Platinum would be too expensive. Industrial chemistry always balances chemistry with economics.

Higher pressure would shift equilibrium more to the right, but at 2 atm the yield is already over 99% at 450°C. Going higher is not cost-effective.

The reaction $\text{SO}_3 + \text{H}_2\text{O} \rightarrow \text{H}_2\text{SO}_4$ is extremely exothermic. Direct contact creates a dense, corrosive acid mist that is nearly impossible to condense or contain. Instead, SO $_3$ is absorbed in concentrated H $_2$ SO $_4$ to form oleum (H $_2$ S $_2$ O $_7$ ), which is then diluted with water safely.

$\text{N}_2 + 3\text{H}_2 \rightleftharpoons 2\text{NH}_3$ , $\Delta H = -92$ kJ/mol

Temperature: Exothermic forward. Lower T shifts right, but rate drops. Compromise at 450°C.
Pressure: 4 mol gas → 2 mol gas. Higher P shifts right. Use 200 atm.
Concentration: Remove NH $_3$ as it forms (by cooling). Shifts equilibrium right. Recycle N $_2$ and H $_2$ .
Catalyst: Fe/Mo does not shift equilibrium but reaches it faster.

Solved Problems (Exam Style)

Problem 1 (NEET pattern): In the Ostwald process, which catalyst is used? (a) Fe (b) V $_2$ O $_5$ (c) Pt/Rh (d) Ni

Ostwald process makes nitric acid from ammonia. The oxidation of NH $_3$ to NO uses a Pt/Rh gauze catalyst at 850°C. Answer: (c)

Quick recall: Fe → Haber, V $_2$ O $_5$ → Contact, Pt/Rh → Ostwald, Ni → hydrogenation.

Problem 2 (CBSE Board): Why is the temperature in the Haber process kept at 450°C even though lower temperature favours equilibrium?

The forward reaction is exothermic, so lower temperature gives a higher equilibrium yield. However, at low temperatures, the rate of reaction becomes impractically slow — it would take too long to reach equilibrium. At 450°C with an iron catalyst, the rate is fast enough for commercial production while still giving a reasonable 15% yield. Unreacted gases are recycled to achieve overall high conversion.

Common Mistakes

Confusing the Haber catalyst (Fe) with Ostwald (Pt) and Contact (V $_2$ O $_5$ ).

Saying Haber process uses high temperature to favour equilibrium. High T actually hurts equilibrium; it is used for kinetics.

Writing that sulphuric acid is made by dissolving SO $_3$ in water directly. It is dissolved in concentrated H $_2$ SO $_4$ first (oleum), then diluted.

Thinking the Solvay process can make K $_2$ CO $_3$ . It cannot, because KHCO $_3$ is too soluble to precipitate. This is a commonly tested edge case.

Forgetting that the Ostwald process depends on the Haber process. The ammonia needed comes from Haber. If a question asks for HNO $_3$ production from scratch, both processes are needed.

Exam Weightage and Revision

This topic is a repeat performer in board papers and entrance exams. NEET typically asks one to two questions on the core mechanisms, CBSE boards give three to six marks, and state PMT papers often include a diagram-based long answer. The PYQs cluster around a small set of facts — lock those and you clear the topic.

NEET 2023 asked about the Contact process catalyst. CBSE 2024 boards had a five-mark question on the Haber process conditions with Le Chatelier’s analysis. JEE Main occasionally tests Born-Haber style industrial thermodynamics. This is a low-effort, high-reward chapter.

When a question gives a scenario, identify the core mechanism first, then match it to the concepts above. Most wrong answers come from reading the scenario too quickly.

Memorise four industrial processes with catalyst and one use each. That table is your whole revision.

Practice Questions

Q1. Why is a promoter used in the Haber process?

Molybdenum (Mo) acts as a promoter — it enhances the activity of the iron catalyst by preventing sintering (aggregation of iron particles at high temperature). A promoter does not catalyse the reaction on its own but makes the actual catalyst more effective and longer-lasting.

Q2. Name three products of the chlor-alkali process.

Chlorine (Cl $_2$ ) at the anode, hydrogen (H $_2$ ) and sodium hydroxide (NaOH) at the cathode. Uses: Cl $_2$ → PVC, water treatment. H $_2$ → ammonia synthesis, fuel. NaOH → soap, paper, textiles.

Q3. What is oleum? Why is it formed as an intermediate?

Oleum is fuming sulphuric acid (H $_2$ S $_2$ O $_7$ ), formed by dissolving SO $_3$ in concentrated H $_2$ SO $_4$ . It is used as an intermediate because dissolving SO $_3$ directly in water produces a dense, corrosive acid mist that is dangerous and wasteful. Absorbing in H $_2$ SO $_4$ is a controlled, safe process.

Q4. In the Solvay process, why does NaHCO $_3$ precipitate out?

NaHCO $_3$ is the least soluble of the four possible salts in the solution (NaCl, NaHCO $_3$ , NH $_4$ Cl, NH $_4$ HCO $_3$ ). When CO $_2$ is bubbled through ammoniacal brine (saturated NaCl + NH $_3$ ), the solution becomes supersaturated with respect to NaHCO $_3$ and it crystallises out.

Q5. Why does the Ostwald process use 850°C instead of a lower temperature?

At 850°C with the Pt/Rh catalyst, the selectivity for NO (desired) over N $_2$ + H $_2$ O (undesired side reaction) is maximised. At lower temperatures, thermodynamics favours the complete oxidation to N $_2$ , which is useless. The high temperature kinetically favours the partial oxidation to NO, which is the product we need.

FAQs

Which is the most produced chemical in the world? Sulphuric acid (H $_2$ SO $_4$ ), with global production exceeding 260 million tonnes per year. Its production is sometimes used as an indicator of a country’s industrial development because it is needed in fertilizers, metal processing, petroleum refining, and chemical synthesis.

Why does the Haber process consume so much energy? The high pressure (200 atm) requires powerful compressors, and the process runs at 450°C continuously. The hydrogen feedstock typically comes from steam reforming of natural gas (CH $_4$ + H $_2$ O → CO + 3H $_2$ ), which is itself energy-intensive. Together, the Haber process accounts for about 1-2% of global energy consumption and about 3% of global CO $_2$ emissions.

Can the Solvay process use KCl instead of NaCl? No. KHCO $_3$ is more soluble than NaHCO $_3$ , so it does not precipitate from the solution. The Solvay process fundamentally depends on the low solubility of NaHCO $_3$ to drive the reaction forward.

What replaced the old lead chamber process for H $_2$ SO $_4$ ? The Contact process replaced the lead chamber process in the early 20th century. The Contact process gives higher concentration acid (up to 98% H $_2$ SO $_4$ vs ~80% from the lead chamber), is more efficient, and produces fewer pollutants.

Industrial chemistry is economics disguised as chemistry. Every condition — temperature, pressure, catalyst — is a cost trade-off.