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Chemistry Y12 · Module 5 · Lesson 8 of 18

IQ2, Le Chatelier's Principle · ★ Consolidation

★ Consolidation, Deepening L05–L07

Consolidation, LCP Mastery

A classic 6-mark HSC question asks you to explain the effect of increasing temperature on both the yield and the rate of ammonia production. Full-mark answers are built the same way every time, from four components, direction, LCP reason, collision theory mechanism, and the effect on Keq. Answers that lose marks usually hold the same knowledge but leave out the activation energy argument.

Optional consolidation. This lesson introduces no new syllabus dot points, it consolidates material taught earlier in the module. It is valuable practice, and it is not required to complete the core route.

No new dot points.

This lesson deepens IQ2 through harder multi-variable LCP problems, graph interpretation, and Band 6 written response practice.

Today's hook: On a 6-mark Haber process question, what separates a full-mark answer is not extra knowledge, it is the four-component structure. Today you build that structure until it's automatic.

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Get oriented

Warm up first

Three quick questions from earlier lessons. Pulling old material back to mind before you learn something new makes the new material stick better, so this is not busywork.

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

Here is the question all three students attempted: "The Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, is at equilibrium. The temperature is increased AND the volume is decreased simultaneously. Predict and explain the effect on the equilibrium position and on Keq."

Student response

"Increasing temperature shifts left (exothermic forward). Decreasing volume shifts right (4 mol → 2 mol). These effects oppose each other, so the net position shift cannot be predicted from Le Chatelier's Principle alone, it depends on the size of each effect. Keq decreases, because temperature is the only variable here that changes Keq."

Student response

"The equilibrium shifts left because of Le Chatelier's Principle. Keq changes because conditions changed."

Student response

"Both changes make the reaction go faster. More ammonia is produced."

Before reading on, rank these answers from best to worst and identify exactly what each student did wrong or right. What specific content did Student B and C miss?

Learning Intentions

Know

Identify precisely what separates full-mark and partial-mark LCP responses, and apply this to your own writing

Understand

Apply a systematic method to multi-variable LCP problems: analyse each disturbance separately, identify reinforcing or opposing effects
Explain why confusing rate with yield is the most consequential IQ2 error, and never make it again

Can Do

Identify any type of disturbance from a concentration-vs-time graph by reading its characteristic signature
Write a 7-mark Band 6 extended response on a multi-variable LCP problem with graph interpretation

Key Terms

Le Chatelier's Principle

A system at equilibrium shifts to partially counteract any applied disturbance.

Concentration shift

Equilibrium shifts away from added species or towards removed species.

Temperature shift

Increase in temperature favours the endothermic direction; ΔH sign determines Keq change.

Pressure/volume shift

Equilibrium shifts towards fewer moles of gas when pressure is increased or volume decreased.

Catalyst effect

Speeds up attainment of equilibrium without changing equilibrium position or Keq.

Keq change

Only temperature changes Keq; concentration, pressure and catalyst changes do not alter Keq.

Core Concepts

Ranking three student answers

Marking Analysis

1. Ranking the Three Students, What Full Marks Requires

Examining a high-scoring, a partial-scoring, and a low-scoring answer to the same question is the fastest way to understand exactly what markers are looking for.

6/6 Student A

Correctly identifies both disturbances and their separate effects. Correctly identifies that they oppose each other, and then does the disciplined thing: it refuses to guess a net position, because when two shifts oppose, the net direction depends on the quantitative size of each effect and cannot be settled qualitatively. It resolves the one part that is unambiguous, Keq decreases, and says specifically why (temperature is the only variable that changes Keq). Demonstrates: separate analysis of each variable, understanding of opposing effects, and correct application of "only temperature changes Keq."

3/6 Student B

Correctly identifies shift left (1 mark). Cites LCP (1 mark). Critical error: "Keq changes because conditions changed", pressure and volume do NOT change Keq. The marker can see the student does not understand which conditions change Keq. No collision theory mechanism, missing 2 marks. No analysis of the volume effect at all, missing 1 mark.

1/6 Student C

"Both changes make the reaction go faster", confuses rate with equilibrium position. "More ammonia is produced", unsupported and almost certainly wrong: neither disturbance is analysed, and the temperature increase on its own shifts an exothermic forward reaction left, reducing ammonia. Student C has committed both the rate/yield confusion AND the temperature direction error, the two most fundamental IQ2 conceptual errors.

The Three Errors That Cost Marks

✗ "Conditions changed, so Keq changes."

✗ Omitting collision theory mechanism from extended responses.

✗ Confusing rate with equilibrium yield.

✓ Only temperature changes Keq. Every other factor leaves Keq unchanged.

✓ Always include: which rate increases more, which activation energy is higher, which direction of shift results.

✓ Rate = how fast equilibrium is reached (kinetics). Yield = proportion of product at equilibrium (thermodynamics). They are independent.

Three errors that cost marks: "conditions changed so Keq changes" (wrong, only temperature changes Keq); omitting collision theory from extended responses; confusing rate (how fast equilibrium is reached) with yield (proportion of product at equilibrium, they are independent, governed by different factors).

Pause, copy the highlighted three errors into your book before moving on.

Which of the following changes will alter the value of Keq for a reaction?

The multi-variable method

Systematic Method

2. Multi-Variable LCP, Systematic Method

We just saw the three errors that most commonly cost marks in LCP extended responses. That raises a question: when a question contains two simultaneous disturbances, what step-by-step method ensures every mark is captured? This card answers it → the six-step multi-variable LCP method.

Multi-variable LCP problems are not harder than single-variable problems, they are two single-variable problems done in sequence, followed by an honest judgement about what can and cannot be resolved when the two effects oppose.

The Six-Step Method

Step 1: Identify every disturbance. List them separately.

Step 2: For each disturbance, determine direction of shift using LCP.

Step 3: For each disturbance, determine effect on Keq (only temperature → Keq changes).

Step 4: Check if shifts reinforce or oppose each other.

Step 5: If reinforcing → state the combined direction. If opposing → state that the net position cannot be determined qualitatively, because the outcome depends on the relative magnitude of each effect, then resolve Keq, which is always decidable.

Step 6: Write the final answer with the position (or a clear statement that the net direction is indeterminate when the effects oppose), the Keq effect, and collision theory for the temperature change.

Example, Haber process, increase T AND decrease volume:

Effect of increasing T: shift left (exothermic forward); Keq decreases
Effect of decreasing volume (increase P): shift right (4 mol → 2 mol); Keq unchanged
Opposing effects, so the net position shift cannot be decided qualitatively, it depends on how large each change is. Net Keq: decreases, and only because the temperature changed.

Always address Keq separately from equilibrium position.

The net position is genuinely undetermined when the effects oppose, but Keq is always unambiguous: it either changes (if T changed) or it doesn't (if T didn't change). That is the markable part, so never leave Keq out of a multi-variable answer.

Common error

Only addressing one disturbance and ignoring the other. Markers expect every disturbance mentioned in the question to be addressed. If the question says "temperature is increased AND volume is decreased," both must be analysed separately before the combined effect is stated.

Six-step multi-variable method: (1) list all disturbances; (2) predict each shift separately; (3) determine each Keq effect; (4) check if reinforcing or opposing; (5) if reinforcing, state the combined direction, if opposing, state that the net position is indeterminate qualitatively and resolve Keq instead (only temperature changes Keq); (6) state the result with collision theory; Keq is always unambiguous even when the position is not.

Add the highlighted six-step method to your notes before continuing.

For 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ/mol, the temperature is decreased AND more O₂ is added. Do these disturbances reinforce or oppose each other?

Disturbance signatures on graphs

Graph Skills

3. Interpreting Concentration-vs-Time Graphs, Disturbance Signatures

We just saw the six-step method for multi-variable LCP problems. That raises a question: how do we identify what type of disturbance occurred when given a concentration-vs-time graph, what is the characteristic visual signature for each? This card answers it → five graph signatures with their immediate and subsequent pattern.

Every type of LCP disturbance leaves a characteristic signature on a concentration-vs-time graph, learning to read these signatures lets you work backwards from the graph to identify what was done to the system.

Adding a reactant

That reactant's concentration line shows a sudden upward jump; all other lines are momentarily unchanged; then reactant decreases (consumed by forward reaction) and products increase; new equilibrium has higher product concentration.

Removing a product

That product's concentration shows a sudden downward drop; all other lines momentarily unchanged; then reactants decrease (forward reaction dominant) and product increases back toward (but not to) original value; new equilibrium has lower reactant concentration.

Increasing temperature (exothermic forward)

No sudden concentration jumps; all product concentrations gradually decrease and all reactant concentrations gradually increase; new horizontal equilibrium values have more reactants and fewer products; Keq has decreased.

Increasing pressure (unequal gas moles)

All gas concentrations suddenly increase simultaneously (volume halved → all concentrations double); then the system shifts toward fewer gas moles, concentrations adjust from this new higher baseline; new equilibrium has more product (if shift right).

Adding a catalyst (to a system already at equilibrium)

NO change in any concentration line, the system is already at equilibrium; both rates increase equally; concentrations remain at the same equilibrium values. The graph appears completely unchanged.

Catalyst signature tested every year

Adding a catalyst to a system already at equilibrium produces NO observable change in a concentration-vs-time graph. If you see a graph where "nothing happens" after an addition, the answer is often that a catalyst was added.

Common error for pressure

When pressure is increased, students often describe only the side that shifted as increasing. In fact, ALL gas concentrations jump up immediately (all concentrations double when volume halves), then the system re-equilibrates from this new higher baseline.

How each disturbance shows on a concentration–time graph, which way equilibrium shifts, and whether Kₑq changes.

Five c-vs-t graph signatures: reactant added → one line sudden ↑; product removed → one line sudden ↓; temperature change → no sudden jump, gradual drift (Keq changes); pressure increase → ALL gas lines sudden ↑ simultaneously; catalyst at equilibrium → no change at all (flat lines remain flat).

Pause, copy the highlighted five graph signatures into your book before the check below.

A concentration-vs-time graph shows that at time t₁, ALL species concentrations (both reactants and products) suddenly increase simultaneously. What disturbance occurred?

Diagnosing a disturbance from a graph

Graph Skills

4. Identifying Disturbances from Graphs, Diagnostic Method

We just saw the five disturbance signatures on c-vs-t graphs. That raises a question: what is the step-by-step method for working backwards from a graph to identify what disturbance was applied? This card answers it → the six-step diagnostic method with the required answer structure.

Being given a graph and asked "what disturbance was applied?" is the reverse of the standard LCP question, and it requires the same knowledge applied backwards.

Six-Step Diagnostic Method

Step 1: Are there any sudden (instantaneous) changes? A sudden change = something added/removed or volume changed. A gradual change (no sudden jump) = temperature change.

Step 2: If sudden: how many species changed? One species → that species was added or removed. All gas species proportionally → volume/pressure change.

Step 3: Identify the subsequent direction of shift, which concentrations increase and which decrease after the sudden change.

Step 4: Match the shift direction to the disturbance using LCP.

Step 5: Check consistency, does the identified disturbance produce the observed shift direction?

Step 6: State the disturbance specifically, including the evidence from the graph.

Answer structure

"At t₁, the concentration of [species] suddenly [increased/decreased], followed by [describe subsequent changes], this indicates [disturbance type] was applied, causing a [left/right] shift." Always describe both the observation AND the explanation.

High-yield study activity

Practice drawing the graph signatures for all five disturbance types from memory, then practice identifying disturbances from graphs you haven't seen. This appears in almost every HSC paper and can be worth 4–6 marks.

Six-step graph diagnostic: (1) sudden vs gradual change?; (2) how many species changed?; (3) direction of subsequent shift; (4) match to LCP disturbance type; (5) verify consistency; (6) state disturbance with graph evidence; answer structure = observation at t₁ + identification of disturbance + shift direction.

Add the highlighted diagnostic method to your notes before the check below.

Gradual re-equilibration follows every disturbance, so it is not by itself diagnostic. True or False: on a concentration-vs-time graph, the absence of a sudden jump at t₁, with all species then drifting gradually to new equilibrium values, is the signature of a temperature change.

Cross-lesson links: The four-component Band 6 structure practised here (direction → LCP → collision theory → Keq) is applied to pressure and catalyst questions in L06 and temperature calculations in L13. The rate/yield confusion in Card 5 is tested every year in HSC exams involving both the Haber process (L07) and Contact Process. ICE tables in L10 give the quantitative version of the yield concept explored here.

The rate and yield confusion

Misconception Resolution

5. The Rate/Yield Confusion, A Dedicated Misconception Resolution

We just saw the six-step graph diagnostic method. That raises a question: what is the single most consequential misconception in Module 5, the one that loses marks in almost every extended response? This card answers it → the rate/yield confusion, with a table separating what controls each.

The most consequential IQ2 error is not a misconception about equilibrium, it is confusing how fast ammonia is produced (rate) with how much ammonia is present at equilibrium (yield). They are governed by completely different factors.

Rate: speed at which the system reaches equilibrium, determined by collision frequency, activation energy, and catalyst.

Yield: proportion of reactants converted to products at equilibrium, determined by Keq and therefore temperature (and pressure for gas-phase reactions with unequal moles).

Change	Effect on Rate	Effect on Yield (Keq)	Notes
Increase T (exothermic fwd)	Increases	Decreases	Classic trade-off
Increase T (endothermic fwd)	Increases	Increases	Both improve
Increase P (more moles reactants)	Increases	Increases	Both improve
Add catalyst	Increases	No change	Rate only
Add reactant	Increases (momentarily)	No change (Keq unchanged)	Position shifts, not Keq

Student C's error was saying "both changes make the reaction go faster", fast is about rate. "More ammonia is produced" requires the yield to increase. For an exothermic reaction, increasing temperature increases rate but DECREASES yield. These are not the same thing and cannot be confused.

Band 6 sentence

"Increasing temperature increases the rate of ammonia production but decreases the equilibrium yield, fewer moles of ammonia are present at equilibrium at the higher temperature." This single sentence demonstrates the rate/yield distinction that separates Band 5 from Band 6.

The exact sentence to identify as wrong

"Using a catalyst improves the yield of the Haber process." Yield is Keq-dependent. Catalyst does not change Keq. Catalyst does not improve yield. The catalyst improves the RATE of reaching equilibrium, not the yield at equilibrium.

Rate (kinetics) ≠ yield (thermodynamics): rate is determined by temperature, catalyst, concentration, and pressure; yield is determined by Keq and therefore temperature (and pressure for unequal gas moles); a catalyst NEVER changes yield; for exothermic forward reactions, increasing temperature increases rate but decreases yield.

Pause, copy the highlighted rate vs yield distinction into your book before the check below.

Rate vs yield. Rate (how fast equilibrium is reached) is increased by higher temperature, a catalyst, higher pressure, higher reactant concentration and greater surface area. Yield (how much is converted, Keq-dependent) is changed only by temperature and, for unequal gas moles, pressure; a catalyst, concentration and surface area do NOT change yield.

A student writes: "Adding an iron catalyst to the Haber process improves the yield of NH₃ at equilibrium." Identify the error in this statement.

✓

Worked examples and activities

Activities

Worked Example 1, Identifying a Disturbance from a Graph

Problem: A concentration-vs-time graph for CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), ΔH = −206 kJ/mol, shows all species at equilibrium. At time t₁: concentrations of CO and H₂ suddenly increase; concentrations of CH₄ and H₂O are momentarily unchanged; then CO and H₂ gradually decrease; CH₄ and H₂O gradually increase to new higher values. (a) Identify the disturbance. (b) Explain why CO and H₂ increased suddenly. (c) Explain the subsequent shift using LCP and collision theory.

(a): Only the reactants (CO and H₂) increased suddenly while products were momentarily unchanged. This is the signature of reactant addition, most likely a feed of synthesis gas (CO + H₂ mixture) was added to the reactor.

(b): Adding CO and H₂ directly increases their concentrations instantaneously. Product concentrations are momentarily unchanged because the equilibrium shift takes time, it requires collisions to occur, which happens over seconds to minutes, not instantly.

(c): LCP: adding reactants (CO and H₂) disturbs equilibrium; system shifts right to partially counteract the disturbance by consuming some of the added reactants. Collision theory: increased concentrations of CO and H₂ increase the frequency of effective forward collisions (more molecules per unit volume → more frequent effective collisions in the forward direction). Forward rate > reverse rate → net forward reaction → CH₄ and H₂O produced → concentrations increase. As CO and H₂ are consumed, forward rate decreases; as CH₄ and H₂O increase, reverse rate increases; equilibrium re-established when rates equalise. Keq is unchanged.

Worked Example 2, Analyse, Multi-Variable LCP with Opposing Effects

Problem: The equilibrium SO₂(g) + NO₂(g) ⇌ SO₃(g) + NO(g), ΔH = −41 kJ/mol, is at equilibrium. An engineer simultaneously decreases the temperature by 100°C and removes some SO₃ product. Note: this reaction has equal moles of gas on both sides (2 mol each), so pressure is irrelevant here. (a) Predict the effect of each disturbance separately. (b) Do the effects reinforce or oppose? (c) State the overall effect on equilibrium position and Keq.

(a, temperature decrease): Forward reaction is exothermic (ΔH = −41 kJ/mol). Decreasing temperature, Le Chatelier shifts in the exothermic direction (forward, right) to release heat and counteract the temperature drop. Shift RIGHT. Keq increases (lower T for exothermic forward reaction → larger Keq → more products at equilibrium).

(a, remove SO₃ product): Removing product decreases its concentration → reverse collision frequency decreases → forward rate > reverse → shift RIGHT. Keq unchanged (concentration change does not affect Keq).

(b): Both disturbances shift the equilibrium to the RIGHT. Effects REINFORCE each other.

(c): Overall equilibrium shift is to the RIGHT. SO₃ and NO concentrations increase; SO₂ and NO₂ concentrations decrease. Keq increases, because temperature decreased for an exothermic forward reaction (concentration change does not affect Keq). The Keq change is due to the temperature decrease only.

Worked Example 3, Evaluate (7 marks), Full Extended Response with Graph Interpretation

Problem: A student monitors the equilibrium N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57 kJ/mol, in a sealed syringe. The syringe contains both N₂O₄ (colourless) and NO₂ (brown) at equilibrium. At t₁ the syringe is compressed to half its original volume; at t₂ the syringe is placed in a hot water bath. (a) Describe and explain the colour change at t₁. (b) Describe and explain the effect of the temperature change at t₂ on equilibrium position, collision theory, and Keq. (c) Explain why a catalyst added at any point would not change the colour of the mixture.

(a, t₁ compression): Immediate observation: the colour darkens suddenly. Compressing to half the volume doubles the concentration of both N₂O₄ and NO₂ instantaneously, more NO₂ molecules per unit volume → more intense brown colour. This is not an equilibrium shift; it is a simple concentration increase. After re-equilibration: the colour becomes paler than the immediate post-compression colour (though still darker than the original). LCP: increasing pressure shifts equilibrium toward fewer moles of gas, left side has 1 mol N₂O₄; right side has 2 mol NO₂. Increase pressure → shift LEFT toward 1 mol (N₂O₄). Collision theory: compression increases concentration of all species → both collision frequencies increase → the reverse reaction (2NO₂ → N₂O₄) involves 2 moles of gas and the forward (N₂O₄ → 2NO₂) involves 1 mole → reverse collision frequency increases proportionally more → reverse rate > forward rate → net reverse reaction → more N₂O₄ forms → brown colour fades from immediate darker colour. Keq: unchanged (pressure does not change Keq).

(b, t₂ temperature increase): Forward reaction is endothermic (ΔH = +57 kJ/mol). Increase temperature → Le Chatelier shifts in endothermic direction (forward, right) to absorb added heat. Shift RIGHT → more NO₂ produced → colour darkens. Collision theory: increasing temperature increases average kinetic energy of all particles. Because the forward reaction is endothermic, it has higher Eₐ(forward) > Eₐ(reverse). A proportionally greater fraction of particles now exceed the forward (higher) activation energy → forward rate increases more than reverse rate → net forward reaction → equilibrium shifts right → more NO₂ → brown colour darkens. Keq increases, temperature change for endothermic forward reaction increases Keq. A new, higher Keq applies at the elevated temperature.

(c, catalyst): A catalyst lowers Eₐ equally for both forward and reverse reactions. At equilibrium, forward and reverse rates were already equal. The catalyst increases both rates by the same factor, they remain equal. No net shift in equilibrium position occurs. The ratio of NO₂ to N₂O₄ at equilibrium is unchanged. Therefore the colour of the mixture is unchanged. Keq is unchanged (only temperature changes Keq).

Interactive Tool, Le Chatelier's Principle Open fullscreen ↗

In the simulator, change the temperature and watch Keq, then change the volume and watch Keq. Now apply the six-step method to the Haber equilibrium (exothermic forward) disturbed by an increase in temperature AND a decrease in volume at the same time. Which statement can you actually defend?

Sort the steps+7 XP

Put the complete LCP analysis method in order for any equilibrium disturbance.

State whether K_eq changes (only temperature changes K_eq).
Write the equilibrium equation and label ΔH direction.
Predict the observable change (yield, colour, pressure).
Identify the specific disturbance applied to the system.
Apply LCP: state the direction of shift that opposes the disturbance.

✓

Practice questions

Multiple Choice

+5 XP

A fresh set drawn from this lesson's question bank, with feedback shown immediately. +5 XP per correct · +25 XP all correct

Extended Response

4 marks

Q6: A concentration-vs-time graph for the Haber process N₂(g) + 3H₂(g) ⇌ 2NH₃(g), ΔH = −92 kJ/mol, shows the system at equilibrium. At time t₁, the concentrations of all four species (N₂, H₂, and NH₃) suddenly and simultaneously increase by the same proportion. The NH₃ concentration then gradually increases further, while N₂ and H₂ concentrations decrease slightly. (a) Identify the disturbance at t₁ and explain why all concentrations increased simultaneously. (b) Explain the subsequent shift using Le Chatelier's Principle and state whether Keq changes.

6 marks

Q7 (Evaluate): The equilibrium 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ/mol, is established in an industrial reactor at 450°C and 1 atm. An industrial chemist simultaneously increases the temperature to 600°C and increases the pressure to 10 atm. Using the six-step multi-variable LCP method, fully analyse the effect of each change separately, identify whether the effects reinforce or oppose, state what can and cannot be concluded about the overall equilibrium position, state the overall effect on Keq, and include a collision theory explanation for the temperature effect.

3 marks

Q8: A student claims that at a higher temperature the iron catalyst in the Haber process becomes more effective and improves both the rate AND the yield of NH₃. Identify every error in this statement and write a fully corrected version.

✓

Model Answers

Show Model Answers

Q6 Model Answer:

(a): The disturbance at t₁ was a decrease in the volume of the container (increase in pressure). The simultaneous and proportional increase in all species' concentrations, including both reactants AND the product, is the signature of volume compression. Simply adding a reactant would only increase that reactant's concentration suddenly; adding N₂ and H₂ would not simultaneously increase NH₃. When volume is halved, all concentrations double instantaneously before any equilibrium shift occurs.

(b): Le Chatelier's Principle: increasing pressure shifts equilibrium toward the side with fewer moles of gas. Left: 1 + 3 = 4 mol gas. Right: 2 mol gas. Equilibrium shifts RIGHT → more NH₃ produced → NH₃ concentration increases further above its doubled value; N₂ and H₂ concentrations decrease slightly below their doubled values. Keq is unchanged, pressure changes do not alter Keq; only temperature does.

Q7 Model Answer (6-mark Band 6):

Step 1, Disturbances: Temperature increase (450°C → 600°C) and pressure increase (1 atm → 10 atm).

Step 2, Temperature effect on position: Forward reaction is exothermic (ΔH = −196 kJ/mol). Increasing temperature favours the endothermic (reverse) direction. Equilibrium shifts LEFT.

Step 3, Temperature effect on Keq: Increasing temperature for an exothermic forward reaction → Keq decreases.

Step 4, Pressure effect on position: Left: 2 + 1 = 3 mol gas. Right: 2 mol gas. Increasing pressure favours fewer gas moles (right). Equilibrium shifts RIGHT.

Step 5, Pressure effect on Keq: Pressure does not change Keq, unchanged.

Step 6, Opposing effects and combined result: Temperature shifts LEFT; pressure shifts RIGHT, so the effects oppose. The net position shift therefore cannot be determined from Le Chatelier's Principle alone: which effect wins depends on the quantitative magnitude of each, which qualitative reasoning cannot supply. State this explicitly rather than guessing a direction. Net Keq: decreases, and only because of the temperature increase (the pressure increase leaves Keq unchanged). Keq is the unambiguous, markable part of this answer.

Collision theory for temperature: Increasing temperature increases the average kinetic energy of all particles. Both forward and reverse rates increase. 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 the higher reverse Eₐ → reverse rate increases more than forward rate → reverse rate > forward rate → net reverse reaction → equilibrium shifts left.

Q8 Model Answer:

Errors: (1) "The catalyst becomes more effective at higher temperature", partially misleading. (2) "Improves the yield of NH₃", wrong. The catalyst does not improve yield. Yield is determined by Keq (and therefore temperature). For the Haber process (exothermic), increasing temperature actually DECREASES yield by decreasing Keq. (3) Implied: that the catalyst causes more NH₃ to be present at equilibrium, wrong.

Corrected statement: "At higher temperatures, the iron catalyst becomes active and allows the reaction to reach equilibrium more quickly (increasing the rate of production). However, the catalyst does not change the equilibrium position or Keq, it does not improve yield. For the exothermic Haber process, increasing temperature actually decreases Keq and therefore decreases the equilibrium yield of NH₃. The catalyst's benefit is kinetic, enabling commercially acceptable reaction rates at the compromise temperature of 400–500°C."

✓

Retrieve and reflect

Check what actually stuck

Extended Response, 8 marks

The equilibrium N₂O₄(g) ⇌ 2NO₂(g), ΔH = +57 kJ/mol, is established in a sealed flask. A student simultaneously compresses the flask to one-third of its original volume AND increases the temperature by 50°C. Using the six-step multi-variable LCP method: (a) analyse each disturbance separately for effect on equilibrium position and Keq; (b) state whether the effects reinforce or oppose; (c) state what can and cannot be concluded about the overall shift, and state the Keq change; (d) write the collision theory explanation for the temperature effect, referencing activation energies. Also describe what would be observed visually (the gas mixture is brown due to NO₂).

How did your thinking change?

Look back at your initial ranking of Student A, B, and C. Recall the 2019 NSW HSC question: students who scored 6/6 included the activation energy argument (the reverse reaction has higher Eₐ, so it increases proportionally more with temperature). Were you correct in your ranking? What specific errors in Student B and C's responses do you now understand more deeply after working through this lesson?

I have completed this lesson

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