S5 Chemistry: Industrial Processes (Paper 2: Intermediate)

2026-11-12

Ten items on compromise conditions and quantitative work: pressure and temperature effects, yield and heat released, and overall conversion with recycling, with answers.

Direct answers　Answers in order: gas moles fall from 4 to 2, exothermic so heating shifts it back, compromise conditions, 15.0 per cent, 60.0 per cent, 46 kJ, towards ammonia with a change of 2 mol, about 57.8 per cent, 184 kJ, about 200 atmospheres is enough.

01　Paper details (check answers afterwards)

Item	Detail
Questions	10
Suggested time	20
Difficulty mix	Foundation 3 · Intermediate 3 · Challenge 3 · Error redo 1
Chapter	Chapter: S5 Chemistry: Industrial Processes
Level	S5

02　Questions (10 items)

FoundationWhy does raising the pressure favour ammonia in the Haber process?
FoundationWhy does too high a temperature lower the equilibrium yield of ammonia?
FoundationWhat is the name given to the set of conditions of about 450 degrees C and 200 atmospheres?
IntermediateA theoretical 4 mol of ammonia is made but only 0.6 mol is obtained. What is the percentage yield?
IntermediateFrom 50 mol of nitrogen the theoretical yield is 100 mol of ammonia, but 60 mol is obtained. What is the yield?
IntermediateThe Haber process has dH = -92 kJ mol-1 for the equation as written. How much heat is released per mole of ammonia?
ChallengeWhen the pressure is raised, which way does the Haber equilibrium move and by how many moles of gas does the count change?
ChallengeIf the single pass conversion is 25 per cent, what is the overall conversion after the unreacted gas is recycled three times?
ChallengeHow much heat is released when 4 mol of ammonia is formed, taking dH as -92 kJ mol-1 for the equation?
Error redoA student argues that because higher pressure always helps, the Haber process should run at 2 000 atmospheres. What is wrong with that?

03　Answers and one-line notes

#	Answer	One-line note
1	The forward direction goes from 4 mol of gas to 2 mol, so higher pressure shifts the equilibrium to the right	N2 + 3H2 -> 2NH3 has 4 mol of gas on the left and 2 mol on the right, so a higher pressure favours the side with fewer gas molecules.
2	Because the forward reaction is exothermic, so heating shifts the equilibrium towards the endothermic reverse direction	For an exothermic reaction, heating always shifts the equilibrium backwards, which follows directly from Le Chatelier principle.
3	Compromise conditions, balancing yield, rate and plant cost together	No single setting makes all three best at once, so an intermediate value is chosen.
4	15.0 per cent	0.6 divided by 4 times 100 gives 15.0, and the unused nitrogen and hydrogen are recycled.
5	60.0 per cent	60 divided by 100 times 100 gives 60.0, and the unreacted gases are returned to the converter.
6	46 kJ	The equation releases 92 kJ while forming 2 mol of ammonia, so each mole accounts for 46 kJ.
7	Towards ammonia, from 4 mol of gas down to 2 mol, a change of 2 mol	Raising the pressure drives the system towards the side with fewer gas molecules so as to reduce the pressure.
8	About 57.8 per cent	One minus 0.75 cubed gives 0.578, which is about 57.8 per cent.
9	184 kJ	Four moles of ammonia is twice the equation as written, so 92 times 2 gives 184 kJ.
10	Plant strength, compression energy and safety all count, and about 200 atmospheres is already enough	The gain in yield soon stops covering the cost of compression and stronger equipment.

04　What to do after marking

After marking, redo only the wrong items and write down which step went wrong (formula, unit, or misread number).
Repeat the same item two days later; move to the next paper only when both attempts are correct.
Parents only need the answer and note columns; there is no need to explain every item.

Explain to a classmate one case where raising the pressure helps but costs too much.Tutor Registration →

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