OxfordAQA International GCSE Chemistry: Factors affecting equilibrium, and production of ammonia and sulfuric acid (9202)

This study guide teaches equilibrium and two industrial processes for OxfordAQA International GCSE Chemistry (9202). It follows the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, and covers every statement in section 3.8.2 Factors affecting equilibrium (a to e) and section 3.8.3 Production of ammonia and sulfuric acid (a to d). No required practical belongs to these two sections. The qualification is untiered, so all of it applies to every candidate, and the specification allows content from any part of the course on either paper.

Quick links: course hub · printable checklist · revision notes for this unit · practice questions for this unit · free diagnostics

What this unit covers

Statement	Your task
3.8.2 a	Say when a reversible reaction reaches equilibrium, and why a closed system is needed
3.8.2 b	Say that the relative amounts of substances at equilibrium depend on the conditions
3.8.2 c	Predict how raising or lowering the temperature changes the yield of the endothermic and exothermic reactions
3.8.2 d	Predict how raising or lowering the pressure changes the yield in a gas reaction, by counting molecules
3.8.2 e	Use these ideas, together with rate, to explain the optimum conditions for an industrial process
3.8.3 a	Name the raw materials for the Haber process and where they come from
3.8.3 b	Explain why ammonia matters for fertilizers and why the world needs fertilizers
3.8.3 c	Describe the Haber process: catalyst, conditions, removing ammonia, recycling
3.8.3 d	Describe the three stages of the contact process and explain the Stage 2 conditions

You will also use reacting-mass calculations, which the quantitative chemistry guide teaches in full.

3.8.2 a and b: what equilibrium means

A reversible reaction can go both ways: reactants form products, and products react to give the reactants back. The symbol ⇌ shows this.

Picture a sealed steel cylinder holding only reactant gases. At first the forward reaction is fast and the backward reaction cannot happen. As products build up, fewer reactant particles remain to collide, so the forward rate drops; meanwhile the backward rate rises. In time the two rates match.

At that point the system is at equilibrium:

• it is a closed system, so nothing can get in or out
• the forward and backward reactions are both still happening
• they happen at exactly the same rate, so the amount of each substance stays constant.

Equal rates does not mean equal amounts. Statement b says what decides the mix: the relative amounts of all the substances at equilibrium depend on the conditions. Change the temperature or pressure and the mixture settles at a new composition.

The word yield in this unit means how much of a product is in the equilibrium mixture. A higher yield of the forward reaction means a larger share of products.

3.8.2 c: changing the temperature

In a reversible reaction, if one direction is exothermic (transfers energy to the surroundings), the other direction is endothermic (takes energy in). The rule in the specification is:

Change	Endothermic reaction	Exothermic reaction
Temperature raised	yield increases	yield decreases
Temperature lowered	yield decreases	yield increases

Raising the temperature favours the direction that absorbs energy.

Worked example 1: methanol from carbon monoxide

CO(g) + 2H₂(g) ⇌ CH₃OH(g), where making methanol is exothermic.

1. Forward = exothermic, so backward = endothermic.
2. Raise the temperature: the yield from the endothermic (backward) reaction increases.
3. So the equilibrium mixture contains less methanol and more carbon monoxide and hydrogen.
4. Lower the temperature: the exothermic (forward) yield increases, so more methanol.

Worked example 2: an endothermic forward reaction

CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g), where the forward direction takes in energy (endothermic).

Raising the temperature increases the yield from the endothermic reaction, here the forward one, so the mixture holds more hydrogen. Always decide which direction is which first.

3.8.2 d: changing the pressure

Pressure only matters for gases. Count the molecules of gas on each side of the balanced symbol equation, using the numbers in front of each formula.

• Increase in pressure favours the reaction that produces the fewest molecules.
• Decrease in pressure favours the reaction that produces the most molecules.

Worked example 3: counting molecules

Equation	Molecules left	Molecules right	Increase in pressure favours
CO(g) + 2H₂(g) ⇌ CH₃OH(g)	1 + 2 = 3	1	forward: more methanol
CH₄(g) + H₂O(g) ⇌ CO(g) + 3H₂(g)	1 + 1 = 2	1 + 3 = 4	backward: less hydrogen
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)	2	2	neither: no change in yield

The last row is a trap. With equal numbers of gas molecules on the two sides, a pressure change alters the rate but not the relative amounts at equilibrium.

Worked example 4: combining both factors

For the methanol reaction (forward exothermic, 3 molecules → 1), the highest equilibrium yield of methanol needs a low temperature and a high pressure.

3.8.2 e: optimum conditions in industry

A plant wants plenty of product per hour, cheaply and safely. Yield alone is not enough.

• A low temperature may give a high yield for an exothermic reaction, but the particles move slowly and collide less often and less energetically, so the reaction is too slow.
• A high pressure may give a high yield and a faster rate, but it needs thick-walled equipment and a lot of energy for compressors, so it is expensive and carries a greater safety risk.
• A catalyst speeds up the forward and backward reactions, so equilibrium is reached sooner. It does not change the yield, but it lets a plant run at a lower temperature than it otherwise could.

The chosen conditions are therefore a compromise between yield, rate, cost and safety. The Haber process is the classic case.

3.8.3 a to c: the Haber process

Raw materials (3.8.3 a)

• Nitrogen comes from the air, which is mostly nitrogen.
• Hydrogen may come from natural gas (methane) or other sources.

The process (3.8.3 c)

In the reactor, the purified nitrogen and hydrogen flow over an iron catalyst. The temperature is about 450 °C and the pressure about 200 atmospheres.

nitrogen + hydrogen ⇌ ammonia

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

Only some of the nitrogen and hydrogen react, because the reaction is reversible and ammonia breaks down again. The gas mixture leaving the reactor is cooled. Ammonia liquefies at a much higher temperature than nitrogen or hydrogen (it boils at about −33 °C, while nitrogen boils at about −196 °C and hydrogen at about −253 °C), so the ammonia turns to liquid and is removed. The unreacted nitrogen and hydrogen are recycled back to the reactor, so the raw materials are not wasted.

Why these conditions? (3.8.2 e applied)

The forward reaction, making ammonia, is exothermic.

Condition	Effect on yield	Effect on rate and cost	Outcome
About 450 °C	Cooler conditions would raise the yield	Cooler conditions would make ammonia far too slowly	Compromise between yield and rate
About 200 atmospheres	4 molecules become 2, so higher pressure increases yield	Higher pressure also increases rate, but costs more and is less safe	Compromise between yield, cost and safety
Iron catalyst	No change	Equilibrium reached faster	Useful product made quickly at 450 °C
Recycling	Not an equilibrium effect	Unreacted gases get many chances to react	Little raw material wasted

Worked example 5: reacting masses in the Haber process

A plant reacts 140 tonnes of nitrogen completely with hydrogen. Calculate the mass of ammonia formed and the mass of hydrogen used. (Aᵣ: N = 14, H = 1)

1. Mᵣ: N₂ = 28, H₂ = 2, NH₃ = 17.
2. Ratio from the equation: 1 N₂ : 3 H₂ : 2 NH₃.
3. Mass ratio: 28 : 6 : 34.
4. Scale by 140 ÷ 28 = 5: hydrogen = 6 × 5 = 30 tonnes; ammonia = 34 × 5 = 170 tonnes.
5. Check: 140 + 30 = 170, so mass is conserved.

In a real plant far less forms per pass, because the reaction is reversible; that is why recycling matters.

3.8.3 b: ammonia and fertilizers

Ammonia is a raw material for fertilizers. It reacts with acids to make ammonium salts, such as ammonium nitrate and ammonium sulfate, which supply nitrogen to crops.

Why does the world need fertilizers?

• Plants need nitrogen to make the proteins they need to grow. Most crops cannot use nitrogen gas from the air directly.
• Each harvest removes nitrogen and other nutrients from the soil, so the soil becomes less fertile year after year.
• The world population keeps growing, but the area of farmland is limited, so more food must come from each field.
• Fertilizers replace the lost nutrients and maximise food yields.

Chain: population → more food per field → nitrogen → ammonia fertilizers.

3.8.3 d: the contact process for sulfuric acid

Sulfuric acid is made industrially by the contact process. It has three stages, and only Stage 2 is the reversible, catalysed step you must explain.

Stage 1: burn sulfur in air to make sulfur dioxide. S(s) + O₂(g) → SO₂(g) (The specification prints this equation with a reversible-reaction arrow.)

Stage 2: sulfur dioxide reacts with more oxygen to make sulfur trioxide. 2SO₂(g) + O₂(g) ⇌ 2SO₃(g) This reaction is exothermic and reversible. It needs a catalyst of vanadium(V) oxide, V₂O₅, a temperature of around 450 °C and atmospheric pressure.

Stage 3: sulfur trioxide reacts with water to make sulfuric acid. H₂O(l) + SO₃(g) → H₂SO₄(aq)

Why these Stage 2 conditions?

• Catalyst (V₂O₅): increases the rate of both directions, so equilibrium is reached quickly. It does not change the yield of sulfur trioxide.
• About 450 °C: making SO₃ is exothermic, so cooler conditions would raise the yield, yet the reaction would crawl. Around 450 °C gives a good yield at a useful speed.
• Atmospheric pressure: 3 molecules of gas become 2, so higher pressure would favour sulfur trioxide. But the equilibrium mixture already contains a high proportion of sulfur trioxide at atmospheric pressure. The small gain in yield is not worth the cost of high-pressure equipment and the energy to compress the gases.

Both processes use about 450 °C as a rate-versus-yield compromise, but only the Haber process needs high pressure for a worthwhile yield.

Worked example 6: mass of acid from sulfur

Calculate the maximum mass of sulfuric acid that can be made from 8.0 tonnes of sulfur. (Aᵣ: S = 32, O = 16, H = 1)

1. Follow sulfur through the stages: 1 S → 1 SO₂ → 1 SO₃ → 1 H₂SO₄. The ratio is 1 : 1.
2. Mᵣ: S = 32, H₂SO₄ = 2 + 32 + 64 = 98.
3. Mass ratio 32 : 98, scaled by 8.0 ÷ 32 = 0.25.
4. Maximum mass of acid = 98 × 0.25 = 24.5 tonnes.

Common errors

• Writing that the reactions stop at equilibrium. Both still happen, at the same rate.
• Claiming that reactant and product amounts match at equilibrium. Each amount stays the same; they need not be equal.
• Applying the temperature rule without first deciding which direction is exothermic.
• Counting atoms instead of molecules when predicting a pressure effect, or forgetting the big numbers in front of formulae.
• Saying a catalyst increases the yield. It increases the rate only.
• Explaining 450 °C with “higher temperature gives more product”. For an exothermic forward reaction it gives less; 450 °C is chosen for rate.
• Giving the contact process a high pressure, or swapping the two catalysts. Haber uses iron at about 200 atmospheres; Stage 2 uses vanadium(V) oxide at atmospheric pressure.

Next steps

Condense this unit with the revision notes, then test yourself with the practice questions. Related: rate of reaction, energy changes and the exam preparation page. The free diagnostics will show which other topics need work.

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards, published by OxfordAQA. This guide covers section 3.8.2 Factors affecting equilibrium and section 3.8.3 Production of ammonia and sulfuric acid.