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O-Level Chemistry · Physical Chemistry

Chemical Equilibrium, Haber Process & Contact Process

Reversible reactions, dynamic equilibrium, Le Chatelier's Principle, the industrial manufacture of ammonia and sulfuric acid — fully covered with worked examples.

⚗️ Pure Chemistry🇸🇬 Singapore O-Level📝 5 practice questionsUpdated May 2026

Contents

  1. Reversible Reactions
  2. Dynamic Equilibrium
  3. Le Chatelier's Principle
  4. Effect of Temperature
  5. Effect of Pressure
  6. Effect of a Catalyst
  7. The Haber Process (Ammonia)
  8. The Contact Process (Sulfuric Acid)
  9. Practice Quiz

1 · Reversible Reactions

Reversible reaction

A reaction in which the products can react to reform the reactants. Shown by a double arrow (⇌) in the equation. Most reactions in closed systems are reversible to some extent.

Example
N₂(g) + 3H₂(g) ⇌ 2NH₃(g)
The forward reaction produces ammonia; the reverse reaction decomposes it back into nitrogen and hydrogen.

In an open system (e.g. products escape), the reaction can go to completion. In a closed system, both forward and reverse reactions occur simultaneously, leading to equilibrium.

2 · Dynamic Equilibrium

Dynamic equilibrium

A state in a closed system where the rate of the forward reaction equals the rate of the reverse reaction, and the concentrations of reactants and products remain constant (though both reactions are still occurring).

Key points about dynamic equilibrium:

⚠️ Common Mistake

Students say "at equilibrium, the concentrations of reactants and products are equal." This is wrong. They are constant, not equal. The equilibrium position can strongly favour products or reactants.

3 · Le Chatelier's Principle

Le Chatelier's Principle

When a system at equilibrium is subjected to a change in conditions (temperature, pressure, concentration), the equilibrium position shifts in the direction that tends to oppose or reduce the effect of that change.

In other words, the system "fights back" against whatever change is imposed on it. This principle lets you predict which way the equilibrium will shift without knowing the full kinetics.

4 · Effect of Temperature

Temperature changes affect the equilibrium position AND the value of the equilibrium constant.

ChangeEffect on equilibrium positionReason
Increase temperatureShifts toward the endothermic directionSystem absorbs the extra heat to oppose the change
Decrease temperatureShifts toward the exothermic directionSystem releases heat to oppose the decrease
✅ Exam Tip

For a temperature change question, always identify whether the forward reaction is exothermic or endothermic first (look for ΔH or "heat" in the equation). Then apply Le Chatelier.

5 · Effect of Pressure (for Gases)

Pressure changes affect equilibria involving gases with different numbers of moles on each side.

ChangeEffect on equilibrium position
Increase pressureShifts toward the side with fewer moles of gas
Decrease pressureShifts toward the side with more moles of gas
Equal moles on both sidesNo effect on equilibrium position

For N₂ + 3H₂ ⇌ 2NH₃: left side has 1+3 = 4 mol gas, right side has 2 mol. Increasing pressure shifts the equilibrium to the right (toward fewer moles), favouring ammonia production.

6 · Effect of a Catalyst

Catalyst and equilibrium

A catalyst increases the rate of both the forward and reverse reactions equally. It does NOT change the equilibrium position or the equilibrium constant — it only helps the system reach equilibrium faster.

⚠️ Very Common Exam Mistake

Students say "adding a catalyst shifts the equilibrium to the right." A catalyst does NOT shift the equilibrium position — it speeds up the approach to equilibrium without changing where the equilibrium lies.

7 · The Haber Process (Manufacture of Ammonia)

Haber Process Equation
N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol
Forward reaction is exothermic. Raw materials: nitrogen (from air) and hydrogen (from natural gas/steam reforming).

Industrial Conditions and Compromise

ConditionValue usedWhy this value?
Temperature~450 °CLower T favours more NH₃ (exothermic), but is too slow. 450°C is a compromise: reasonable yield at acceptable speed.
Pressure~200 atmHigher P favours NH₃ (fewer moles). Very high pressure is costly and dangerous; 200 atm is economically viable.
CatalystIron (with promoters)Speeds up the reaction without changing equilibrium position; allows lower temperature to be used.
RecyclingUnreacted N₂ and H₂ recycledIncreases overall yield and efficiency; prevents waste.
📝 Exam-Style Answer: Why is 450°C used rather than a lower temperature?

Although a lower temperature would increase the equilibrium yield of ammonia (forward reaction is exothermic), the rate of reaction would be too slow to be economically viable. 450°C represents a compromise between an acceptable yield and a sufficiently fast rate of reaction.

8 · The Contact Process (Manufacture of Sulfuric Acid)

Sulfuric acid (H₂SO₄) is one of the most widely produced industrial chemicals. The Contact Process has three key stages:

Stage 1 — Burning sulfur
S(s) + O₂(g) → SO₂(g)
Stage 2 — Oxidising SO₂ (the equilibrium step)
2SO₂(g) + O₂(g) ⇌ 2SO₃(g) ΔH = −196 kJ/mol
Conditions: ~450°C, 1–2 atm, vanadium(V) oxide (V₂O₅) catalyst
Stage 3 — Forming sulfuric acid
SO₃ + H₂SO₄ → H₂S₂O₇ (oleum) → H₂SO₄ (diluted with water)
SO₃ is not dissolved directly in water (too violent, creates acid mist). It is dissolved in concentrated H₂SO₄ to form oleum, which is then carefully diluted.
⚠️ Exam Trap

Students often write that SO₃ is dissolved directly in water. This is wrong and dangerous — the process uses concentrated sulfuric acid to absorb SO₃, producing oleum, which is then diluted.

🧠 Quick Practice Quiz

8 questions · O-Level style · Instant feedback

Question 1 of 8
For the reaction: A(g) + 2B(g) ⇌ C(g) + D(g) ΔH = +60 kJ/mol
What happens to the equilibrium position when temperature is increased?
ΔH = +60 kJ/mol means the forward reaction is endothermic. By Le Chatelier's Principle, increasing temperature shifts equilibrium toward the endothermic direction — to the right, producing more C and D.
Question 2 of 8
In the Haber Process, why is a temperature of about 450°C used rather than a much lower temperature?
Lower temperatures favour more NH₃ (exothermic forward reaction) but the rate is too slow. 450°C is a compromise — reasonable equilibrium yield at an economically acceptable rate.
Question 3 of 8
What effect does adding a catalyst have on a system at equilibrium?
A catalyst increases the rate of both forward and reverse reactions equally, so the equilibrium position is unchanged. It only helps the system reach equilibrium faster.
Question 4 of 8
In the Contact Process, why is SO₃ not dissolved directly in water to make H₂SO₄?
Dissolving SO₃ directly in water is extremely exothermic and creates a dense acid mist that is hard to contain. Instead, SO₃ is absorbed into concentrated H₂SO₄ to form oleum (H₂S₂O₇), which is then carefully diluted with water.
Question 5 of 8
For the equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g)
What happens to the equilibrium position when pressure is increased?
Left side: 2 + 1 = 3 mol gas. Right side: 2 mol gas. Increasing pressure shifts equilibrium toward fewer moles of gas — to the right, producing more SO₃.
Question 6 of 8
For the reaction N₂(g) + 3H₂(g) ⇌ 2NH₃(g) ΔH = −92 kJ/mol, what is the effect of decreasing temperature?
Decreasing temperature removes heat from the system. By Le Chatelier's Principle, the equilibrium shifts in the direction that produces heat — the exothermic forward reaction — producing more NH₃. This increases yield but slows the rate, which is why a compromise temperature is used industrially.
Question 7 of 8
What is the role of the iron catalyst in the Haber Process?
A catalyst lowers activation energy for both forward and reverse reactions equally. It speeds up attainment of equilibrium without changing the equilibrium position or yield. In the Haber Process, the iron catalyst allows a lower operating temperature to be used economically.
Question 8 of 8
In the Contact Process, vanadium(V) oxide (V₂O₅) is used as a catalyst in Stage 2. At which temperature range is it used?
The Contact Process Stage 2 (2SO₂ + O₂ ⇌ 2SO₃) uses V₂O₅ catalyst at 450–500°C. Lower temperature would increase yield (exothermic) but the rate would be too slow; higher temperature reduces yield. This is the same compromise principle as the Haber Process.
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