🏭

Chemistry Y12 · Module 5 · Lesson 7 of 18

IQ2 — Le Chatelier's Principle

Industrial Applications & Collision Theory Explanations

In 1913, BASF's Carl Bosch scaled Fritz Haber's laboratory process to industrial production at Oppau, Germany — operating at 450°C, 200 atm, and using an iron catalyst promoted with K₂O and Al₂O₃. The equilibrium yield per pass was only 15%, but continuous recycling raised overall efficiency to over 90%.

Analyse Evaluate Justify

Today's hook: In 1913, Carl Bosch ran the Haber process at 450°C and 200 atm at Oppau — conditions that cut equilibrium yield to 15% per pass but produced ammonia fast enough to be commercially viable. Why does industry accept a lower yield?

0/5TASKS

Worksheets

Practise this lesson

Four printable worksheets that build from the foundations up to exam-style questions — start at whatever level suits you.

Build Foundations & guided practice Apply Application practice Master Mastery challenge Exam Exam-style questions

Think First

You are the chief engineer for an ammonia plant. Your goal is to maximise profit — which means maximising both the yield of ammonia per pass AND the rate at which ammonia is produced. You know: N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol. From Le Chatelier's Principle (L05–L06), you know that low temperature increases yield (exothermic forward reaction) but decreases rate; high pressure increases yield (4 mol gas → 2 mol gas) and increases rate; iron catalyst increases rate but does not affect yield. Write what temperature and pressure you would choose if you cared only about yield. Then write what you would choose if you cared only about rate. The gap between your two answers is the industrial compromise this lesson explains.

Learning Intentions

Know

Explain every LCP observation using collision theory — not just direction, but mechanism
Read and interpret yield-vs-temperature and yield-vs-pressure graphs for industrial equilibrium data

Understand

Analyse the effect of temperature, pressure, and catalyst on both yield and rate of the Haber process
Justify the industrial compromise conditions (400–500°C, 150–300 atm, iron catalyst, recycling) with reference to rate–yield trade-offs

Can Do

Write Band 6 LCP responses using the four-component structure: direction → LCP reason → collision theory → new equilibrium

Key Terms

Haber process

Industrial synthesis of ammonia: N₂ + 3H₂ ⇌ 2NH₃; run at ~450°C, 150–300 atm, iron catalyst.

Contact process

Industrial synthesis of sulfuric acid via SO₂ + ½O₂ ⇌ SO₃; uses vanadium(V) oxide catalyst.

Industrial compromise

The optimum conditions that balance reaction rate and equilibrium yield in a cost-effective way.

Heterogeneous catalyst

A catalyst in a different phase from the reactants (e.g., solid iron catalyst with gas-phase reactants).

Yield

The amount of product obtained; in industrial processes a compromise between thermodynamic yield and reaction rate.

Collision frequency

The number of effective collisions per unit time; increased by higher temperature, pressure, and concentration.

Core Concepts

Analyse

1. Collision Theory Explanations for All LCP Disturbances

Le Chatelier's Principle tells you what happens — collision theory tells you why it happens. Band 6 answers always contain both.

Every LCP prediction has a collision theory explanation. The connection is: a disturbance upsets the balance between forward and reverse collision frequencies; the system shifts in the direction of the faster reaction until rates re-equalise.

Concentration increase (add reactant): more reactant particles per unit volume → more frequent reactant-reactant collisions → forward collision frequency increases → forward rate > reverse rate → shift right until rates re-equalise
Temperature increase (exothermic forward reaction): all particles gain kinetic energy → both forward and reverse rates increase → but the reverse reaction has higher Eₐ → a proportionally greater fraction of particles now exceed the reverse Eₐ → reverse rate increases more → reverse rate > forward rate → shift left until re-equalise
Pressure increase (more gas moles on left): all gas concentrations increase → both collision frequencies increase → the forward reaction involves more gas molecules per reaction event → forward collision frequency increases more → forward rate > reverse rate → shift right
Catalyst: lowers Eₐ equally for both → both rates increase by same factor → rates remain equal → no shift

Disturbance	Why Forward Rate Changes	Why Reverse Rate Changes	Net Effect	Shift
Add reactant	Increases (more particles)	Unchanged initially	Forward > reverse	Right →
Remove product	Unchanged initially	Decreases (less product)	Forward > reverse	Right →
Increase T (exothermic fwd)	Increases	Increases MORE (higher Eₐ)	Reverse > forward	Left ←
Increase P (more moles left)	Increases more (more mol gas)	Increases less	Forward > reverse	Right →
Catalyst	Increases	Increases equally	No change	No shift

Band 6 requirement

In every HSC extended response about LCP, include both the LCP prediction AND the collision theory mechanism. "Increasing temperature shifts the equilibrium left because Le Chatelier's Principle states the system opposes the added heat" is worth 2 marks. Adding the collision theory mechanism (activation energies, relative rate increase) is worth 4–5 marks.

Common error

"Higher temperature makes particles move faster and react more." This does not explain why the equilibrium shifts in a particular direction. The key is the RELATIVE increase in forward vs reverse rates, which depends on which direction has the higher activation energy. Always specify which direction has the higher Eₐ.

Collision theory explains every LCP shift: concentration increase → more frequent collisions in that direction; temperature increase → the direction with higher Eₐ sees a proportionally greater rate increase; pressure increase → side with more gas moles has greater collision frequency increase; catalyst → lowers Eₐ equally, no shift.

Pause — copy the highlighted collision-theory mechanism table into your book before moving on.

A catalyst is added to a reaction at equilibrium. What happens to the forward and reverse rates?

Analyse

2. The Haber Process — Applying All Variables Simultaneously

We just saw how collision theory explains every LCP disturbance. That raises a question: what is the real-world system where ALL four variables — concentration, temperature, pressure, and catalyst — must be balanced simultaneously? This card answers it → the Haber process variable trade-off table.

The Haber process is the ultimate test of IQ2 knowledge — every variable from LCP (concentration, temperature, pressure, catalyst) applies simultaneously, and optimising one variable always costs something in another.

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

Gas moles: left = 4 (1 N₂ + 3 H₂) | right = 2 (2 NH₃)

Variable	Effect on Yield	Effect on Rate	Trade-off?
Low temperature	Increases (Keq larger)	Decreases (slower kinetics)	YES — yield vs rate
High temperature	Decreases (Keq smaller)	Increases (faster kinetics)	YES — rate vs yield
High pressure	Increases (LCP shift right)	Increases (higher concentration)	NO — both improve
Iron catalyst	No change	Increases significantly	NO — rate benefit only

Pressure has no trade-off

Pressure improves both yield AND rate for the Haber process. The only reason industrial plants do not use extremely high pressures (e.g. 1000+ atm) is engineering cost and safety — the vessels, pumps, and seals required become exponentially more expensive and dangerous. This is an economic constraint, not a chemical one.

Insight

At room temperature (25°C), Keq for the Haber process is approximately 6 × 10⁵ — thermodynamically, almost all N₂ and H₂ would convert to NH₃. But the rate at 25°C without a catalyst is essentially zero — the activation energy barrier is too high. This is the fundamental dilemma: thermodynamically ideal conditions are kinetically useless.

N₂ + 3H₂ ⇌ 2NH₃, ΔH = −92 kJ/mol: temperature has a yield-vs-rate trade-off (only variable with this); pressure improves both yield and rate (no trade-off, limited by engineering cost); iron catalyst improves rate only; recycling unreacted gases achieves >95% overall conversion from 15% per-pass yield.

Add the highlighted Haber process variable summary to your notes before continuing.

True or False: Increasing pressure in the Haber process improves yield but decreases the rate of ammonia production.

Haber process mind map — all four industrial variables and their trade-offs

Evaluate

3. The Industrial Compromise — Why 400–500°C, 150–300 atm, Iron Catalyst

We just saw the Haber process variable trade-off table. That raises a question: why were the specific industrial conditions of 400–500°C, 150–300 atm, and iron catalyst chosen — what are the justifications for each? This card answers it → with the reasoning behind every industrial compromise condition.

Temperature — 400–500°C: A compromise between yield and rate. At 400°C, only about 15% conversion per pass at 200 atm — not thermodynamically ideal (room temperature would give higher Keq) but provides commercially acceptable rate. Lower temperatures (e.g. 200°C) would give higher equilibrium yield but the rate would be too slow — the catalyst is inactive below approximately 300°C.

Pressure — 150–300 atm: Improves both yield and rate. At 200 atm and 450°C, approximately 15–25% conversion per pass. Increasing to 1000 atm would improve conversion to ~40% but engineering costs for high-pressure vessels are enormous and safety risks significant. The economics favour moderate pressure with recycling.

Catalyst — iron (with K₂O and Al₂O₃ promoters): Identified by Alwin Mittasch after over 6500 experiments. Allows commercially acceptable rates at 400–500°C. Without the catalyst, ~600–700°C would be needed for acceptable rates — but at these temperatures Keq is so small that the process would not be viable. The catalyst is what makes the entire industrial compromise possible.

Recycling — of unreacted N₂ and H₂: Unreacted gases (80–85% of feed gas per pass) are separated from the NH₃ product by condensation and recycled. This gives overall conversion of >95% despite the low per-pass yield.

HSC extended response

Always address all four factors — temperature (compromise), pressure (engineering cost), catalyst (rate only), and recycling (compensates for low per-pass yield). Missing any one factor costs marks.

Common error

"The catalyst improves the yield of the Haber process." It does not. The catalyst improves the RATE — allowing equilibrium to be reached quickly at 400–500°C. The yield at equilibrium is determined by Keq (temperature and pressure), not by the catalyst.

Industrial compromise: 400–500°C (rate-yield compromise — lower T gives higher Keq but too slow; catalyst inactive below ~300°C); 150–300 atm (engineering cost limit, not chemical limit); iron catalyst (enables acceptable rate at compromise T); recycling of unreacted N₂/H₂ achieves >95% overall yield.

Pause — copy the highlighted industrial conditions and their justifications into your book before the check below.

Why doesn't the Haber process use a much higher pressure, such as 1000 atm, to get an even better yield and rate?

Justify

4. Writing Full Collision Theory Justifications — Band 6 Structure

We just saw why specific industrial conditions were chosen for the Haber process. That raises a question: what exactly is the difference between a Band 4 and a Band 6 HSC answer for LCP questions — and what structure do we follow? This card answers it → the four-component Band 6 response structure for any LCP disturbance.

The difference between a Band 4 and a Band 6 answer for a Le Chatelier question is not knowledge — it is the precision and structure of the explanation.

Four-Component Band 6 Structure

Component 1 — State the direction of shift: "The equilibrium shifts to the right (forward direction)."

Component 2 — State the LCP reason: "This is because Le Chatelier's Principle states the system shifts to oppose the disturbance — adding reactant is opposed by consuming some of the added reactant."

Component 3 — Collision theory mechanism: "Adding reactant increases its concentration, increasing the frequency of effective collisions in the forward direction. The forward reaction rate now exceeds the reverse reaction rate."

Component 4 — New equilibrium description: "The system shifts right until the forward and reverse rates re-equalise at a new equilibrium position with higher product concentration."

For temperature effects, Component 3 must reference activation energies:

"Increasing temperature increases the average kinetic energy of particles. Because the reverse reaction has a higher activation energy (Eₐ reverse > Eₐ forward for an exothermic forward reaction), a greater proportion of particles can now exceed the reverse activation energy. The reverse rate increases more than the forward rate."

Practice until automatic

Practice writing all four components for each type of disturbance — concentration, temperature, and pressure — until the structure is automatic. Students who have practised this structure score full marks; students who know the content but write it in the wrong order or miss components do not.

Why collision theory matters

LCP alone is phenomenological — it describes what happens without explaining why. Collision theory provides the mechanistic explanation. The HSC rewards mechanistic understanding because it demonstrates you can reason about new situations, not just recall memorised predictions.

Band 6 LCP structure — four components: (1) state direction of shift; (2) LCP reason (system opposes the disturbance); (3) collision theory mechanism (which rate changes and why); (4) new equilibrium description; for temperature, component 3 must specify which direction has higher Eₐ.

Add the highlighted four-component structure to your notes before the check below.

In a Band 6 LCP response about a temperature increase on an exothermic reaction, which component is most commonly missing from partial-mark answers?

Cross-lesson links: Carl Bosch's 1913 Oppau conditions introduced in the hook are the standard exam context for both the Haber process (IQ2) and Contact Process questions. The rate/yield compromise analysis in Cards 1–4 underpins L08 (consolidation — LCP mastery). Graph interpretation skills in Card 5 are directly tested in L10 (ICE table calculations) and L12 (Q prediction).

Analyse

5. Graph Interpretation — Reading Industrial Equilibrium Data

We just saw the four-component Band 6 structure for LCP responses. That raises a question: how do we read industrial equilibrium data from graphs — what are the visual signatures of each type of disturbance? This card answers it → with three graph types and the pattern to identify each disturbance.

Industrial chemistry data is almost always presented as concentration-vs-time or yield-vs-condition graphs. Being able to extract LCP information from these graphs is a core HSC skill tested in every Module 5 exam.

Type 1 — Concentration-vs-time with labelled disturbances:

If all concentrations suddenly increase simultaneously → likely pressure increase (volume decrease)
If one reactant suddenly jumps while products unchanged → likely addition of that reactant
If product concentrations decrease and reactant concentrations increase gradually (no sudden jump) → likely temperature increase for exothermic forward reaction
If nothing changes → catalyst was added (to a system already at equilibrium)

Type 2 — Yield-vs-temperature:

For exothermic forward reaction: yield decreases as temperature increases (Keq decreases)
The graph shows a downward curve — high yield at low T, low yield at high T
Industrial operating temperature is never at the yield maximum (lowest T) because rate would be too slow

Type 3 — Yield-vs-pressure:

For reaction with more gas moles on the reactant side: yield increases with pressure
A smooth upward curve; industrial pressure chosen where yield improvement no longer justifies engineering cost

Graph interpretation answer structure

When asked to identify a disturbance from a graph, describe BOTH the immediate change (what happened at the moment of disturbance) AND the subsequent change (how the system re-established equilibrium). "At t₁, the concentration of NH₃ suddenly decreased and then gradually increased back toward (but not reaching) its original value, while N₂ and H₂ concentrations increased — this is consistent with removal of NH₃ product, which caused the system to shift right."

Common error

Only identifying the disturbance from the species whose concentration changed most dramatically. Always check ALL species and their relative changes — the complete pattern identifies the disturbance type unambiguously.

Graph signatures: sudden jump in one species = addition of that species; all concentrations jump simultaneously = pressure increase; gradual drift to new equilibrium (no sudden jump) = temperature change; no change at all = catalyst added to system already at equilibrium; always describe both immediate change AND subsequent re-equilibration.

Pause — write the highlighted graph signature rules into your book before the check below.

A concentration-vs-time graph shows no sudden change — all species' concentrations gradually drift to new equilibrium values (products decrease, reactants increase). What type of disturbance most likely occurred?

Activities

Haber Process Summary

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

Effect on yield:

Lower T → higher yield (Keq increases as T decreases for exothermic forward)
Higher P → higher yield (4 mol gas → 2 mol gas, LCP shifts right)

Effect on rate:

Higher T → faster rate (more particles exceed Eₐ)
Higher P → higher concentration → more frequent collisions → faster rate
Catalyst → increases rate without changing yield (Keq unchanged)

Industrial compromise: 400–500°C, 150–300 atm, iron catalyst, recycling unreacted gases (>95% overall conversion)

Worked Example 1 — Band 5–6 (5 marks)

Problem: The reaction 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ/mol, is at equilibrium. The temperature is increased from 450°C to 550°C. Write a complete Band 6 explanation of the effect on the equilibrium position, the concentrations of all species, and the value of Keq.

Component 1 — Direction: The equilibrium shifts to the LEFT (reverse direction).

Component 2 — LCP reason: The forward reaction is exothermic (ΔH = −196 kJ/mol). Le Chatelier's Principle states the system shifts to oppose the disturbance — increasing temperature is opposed by shifting in the endothermic direction (reverse), which absorbs heat.

Component 3 — Collision theory: Increasing temperature increases the average kinetic energy of all particles. Both forward and reverse reaction rates increase. However, because the reverse reaction is endothermic, it has a higher activation energy (Eₐ reverse > Eₐ forward for an exothermic forward reaction). A proportionally greater fraction of particles now have sufficient energy to exceed the higher reverse Eₐ. The reverse reaction rate increases more than the forward rate → reverse rate > forward rate → net reverse reaction.

Component 4 — New equilibrium and Keq: The system shifts left until rates re-equalise. At the new equilibrium: [SO₃] decreases; [SO₂] and [O₂] increase. Keq = [SO₃]²/([SO₂]²[O₂]) is smaller at the new equilibrium (lower numerator, larger denominator) → Keq decreases. A new, lower Keq applies at 550°C.

Worked Example 2 — Band 6 (6 marks)

Problem: Analyse the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, explaining why each industrial condition is used: (a) temperature of 400–500°C; (b) pressure of 150–300 atm; (c) iron catalyst; (d) recycling of unreacted gases.

(a) Temperature 400–500°C: Low temperature gives high yield (Keq larger for exothermic reaction at lower T) but unacceptably slow rate — the activation energy is too high for sufficient effective collisions, even with catalyst. High temperature gives fast rate but low yield (Keq smaller — equilibrium shifts left). 400–500°C is a compromise — the iron catalyst becomes active at ~400°C, providing acceptable rate at a temperature where Keq still gives meaningful yield of ~15–25% per pass.

(b) Pressure 150–300 atm: High pressure shifts equilibrium right (4 mol gas left → 2 mol gas right, LCP) AND increases collision frequency (higher concentration) → both yield and rate improve. 150–300 atm is a compromise — higher pressures (e.g. 1000 atm) would improve yield further but require enormously expensive high-pressure engineering beyond the commercial value of the additional ammonia produced.

(c) Iron catalyst: The iron catalyst lowers Eₐ for both forward and reverse reactions equally — it does not shift equilibrium or change Keq, so it does not improve yield. Its value is to allow the reaction to reach equilibrium quickly at 400–500°C. Without the catalyst, ~600–700°C would be needed for acceptable rates — but at these temperatures Keq is so small that yield would be commercially unviable.

(d) Recycling: Per-pass conversion at industrial conditions is only ~15–25%. Unconverted N₂ and H₂ are separated from the NH₃ product (by condensation — NH₃ liquefies at higher temperatures than N₂ and H₂) and recycled back into the reactor feed. This allows overall conversion of over 95% despite the low per-pass yield, making the process commercially viable.

Interactive Tool — Le Chatelier's Principle Open fullscreen ↗

Use the Le Châtelier simulator. Adding more REACTANT to a system at equilibrium shifts the equilibrium position…

Fill the blanks+4 XP

Complete these statements about the Haber process conditions and the trade-offs involved.

The Haber process reaction: N&sub2;(g) + 3H&sub2;(g) ⇌ 2NH&sub3;(g) ΔH = −92 kJ mol¹

A lower temperature favours (exothermic forward reaction), but gives a slower .

The industrial compromise temperature (~450 °C) prioritises reaction over maximum yield.

A catalyst (iron) increases the rate but does NOT change .

Complete the Learn phase to unlock practice questions.

✓

Checkpoint Questions

1 mark

Q1: Which correctly explains, using collision theory, why increasing temperature shifts the Haber process equilibrium to the LEFT?

A) Higher temperature increases collision frequency, which favours the forward reaction producing more NH₃

B) Higher temperature increases the average kinetic energy — because the reverse reaction has a higher activation energy, the reverse rate increases proportionally more than the forward rate, shifting equilibrium left

C) Higher temperature always shifts equilibrium in the endothermic direction because heat is a reactant in endothermic reactions

D) Higher temperature reduces Keq, which directly causes the equilibrium to shift left without changing reaction rates

1 mark

Q2: An industrial plant operates the Haber process at 450°C and 200 atm with an iron catalyst. A process engineer proposes increasing the pressure to 400 atm. Which correctly evaluates this proposal?

A) The proposal should be rejected because increasing pressure decreases Keq, reducing yield

B) The proposal would increase both yield and rate but may not be economically justified due to increased engineering costs for high-pressure equipment

C) The proposal would increase yield but decrease rate because higher pressure reduces collision frequency

D) The proposal would increase rate but decrease yield because pressure shifts the equilibrium left for the Haber process

1 mark

Q3: A chemist removes ammonia from the Haber process reactor as it is produced, maintaining a low NH₃ concentration. Using Le Chatelier's Principle and collision theory, which effect does this have?

A) No effect — removing product does not disturb equilibrium because it is a product, not a reactant

B) Keq increases because the ammonia concentration is lower, changing the equilibrium constant

C) Equilibrium continuously shifts right as NH₃ is removed — the reverse rate decreases (less NH₃) while forward rate remains higher — increasing overall yield of NH₃ over time

D) Equilibrium shifts left because removing product decreases the total pressure, favouring the side with more gas moles

1 mark

Q4: For the Contact Process 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ/mol, the vanadium(V) oxide catalyst is used. What is the primary reason for using this catalyst?

A) It allows the reaction to reach equilibrium more quickly at a lower, safer temperature, improving the rate without changing the equilibrium yield

B) It shifts the equilibrium position to the right, increasing the yield of SO₃

C) It increases the value of Keq, making the forward reaction more favourable

D) It prevents the reverse reaction from occurring, ensuring complete conversion to SO₃

1 mark

Q5: At 25°C, Keq for N₂(g) + 3H₂(g) ⇌ 2NH₃(g) is approximately 6 × 10⁵. Why is the industrial process NOT run at 25°C?

A) Keq is too small at 25°C, meaning the yield would be very low

B) The equilibrium shifts left at low temperatures

C) The iron catalyst only works at temperatures above 600°C

D) The rate at 25°C is essentially zero — activation energy cannot be overcome — so equilibrium is never reached in a practical timeframe

Short Answer

5 marks

Q6: Using the four-component Band 6 structure, write a complete explanation of the effect of increasing temperature on the Contact Process equilibrium: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ/mol. Your answer must include: direction of shift; LCP reasoning; collision theory mechanism with reference to activation energies; effect on concentrations; and effect on Keq.

4 marks

Q7: A concentration-vs-time graph for the Haber process shows that at time t₁, the concentrations of N₂ and H₂ suddenly increase significantly while the concentration of NH₃ is momentarily unchanged. The concentrations then gradually change toward new equilibrium values. (a) Identify the most likely disturbance at t₁. (b) Describe what happens to all concentrations after t₁ and explain using LCP and collision theory.

✓

Model Answers

Show Model Answers

Q6 Model Answer (5-mark Band 6):

Component 1 — Direction: The equilibrium shifts to the LEFT (reverse direction).

Component 2 — LCP reason: The forward reaction is exothermic (ΔH = −196 kJ/mol). Le Chatelier's Principle states the system shifts in the direction that minimises the effect of the disturbance — increasing temperature adds thermal energy, which is minimised by shifting in the endothermic (reverse) direction, which absorbs heat.

Component 3 — Collision theory: Increasing temperature increases the average kinetic energy of all particles. Both the forward and reverse reaction rates increase. However, the reverse (endothermic) reaction has a higher activation energy than the forward (exothermic) reaction. A proportionally greater fraction of particles now have sufficient kinetic energy to exceed this higher reverse activation energy. The reverse reaction rate increases more than the forward rate → reverse rate > forward rate → net reverse reaction.

Component 4 — Concentrations and Keq: [SO₃] decreases; [SO₂] and [O₂] increase as the system shifts left. The ratio [SO₃]²/([SO₂]²[O₂]) is smaller at the new equilibrium → Keq decreases. A new, lower Keq applies at the higher temperature. Temperature is the only variable that changes Keq.

Q7 Model Answer:

(a): The disturbance at t₁ was the addition of N₂ and H₂ (reactants) to the system — possibly as a new batch of synthesis gas. The sudden simultaneous increase in both reactant concentrations with unchanged product concentration is the signature of reactant addition, not pressure change (which would increase all species simultaneously including NH₃).

(b): After t₁: N₂ and H₂ concentrations gradually decrease toward their new equilibrium values (some is consumed); NH₃ concentration gradually increases to a new, higher equilibrium value (more product formed). LCP: adding reactants (N₂ and H₂) disturbs equilibrium; system shifts right to minimise the disturbance by consuming some of the added reactants. Collision theory: increased concentrations of N₂ and H₂ increase the frequency of effective forward collisions → forward rate > reverse rate → net forward reaction → NH₃ produced → concentrations change until rates re-equalise at new equilibrium. Keq is unchanged (concentration change does not affect Keq).

Extended Response

An industrial chemist is designing a new plant to produce ammonia using the Haber process. They are considering three changes to the standard conditions: (1) lowering the temperature from 450°C to 350°C; (2) increasing the pressure from 200 atm to 400 atm; (3) adding a more active catalyst that functions at 350°C. For each proposed change, evaluate the effect on both the rate and yield of ammonia production, with reference to collision theory and Le Chatelier's Principle. Then recommend the overall set of conditions and justify your recommendation.