Year 12 Chemistry Module 5 ⏱ ~35 min Lesson 3 of 18

Collision Theory Applied to Equilibrium

A reversible reaction never really stops. Collision theory tells you why the forward rate falls, why the reverse rate climbs, and why the two must eventually meet at a value that is equal but never zero.

Today's hook, In 1884, Henry Louis Le Chatelier cooled a sealed tube of brown NO₂ gas in Paris and watched the colour fade without removing a single molecule. Collision theory is the only thing that explains it.

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Think First

In 1884, Henry Louis Le Chatelier at the École des Mines in Paris cooled a sealed tube of N₂O₄/NO₂ mixture from 25°C to 0°C and recorded that the brown colour faded, even though he hadn't removed any gas. He needed collision theory to explain why. A sealed container holds N₂O₄(g) (colourless) and NO₂(g) (brown) at equilibrium at 25°C. The mixture is a pale brown colour, both species are present.

The container is now placed in an ice bath. Before reading any theory, predict what happens to the colour over the next few minutes. Does the brown get darker, lighter, or stay the same? Explain your reasoning using what you know about how particles collide. Write your prediction and reasoning now. You will revisit this at the end of the lesson with a full collision theory explanation.

Key Relationships, This Lesson

Rate ∝ effective collision frequency

Effective collision: correct orientation + energy ≥ Eₐ
Forward rate ∝ [reactants], decreases as reactants consumed
Reverse rate ∝ [products], increases as products accumulate

ΔH = Eₐ(forward) − Eₐ(reverse)

Exothermic forward: Eₐ(fwd) < Eₐ(rev) → products lower energy
Endothermic forward: Eₐ(fwd) > Eₐ(rev) → products higher energy

Catalyst: lowers Eₐ equally for both directions → no change to equilibrium position or Keq

Learning Intentions

Know

How collision theory explains the approach to equilibrium
The relationship ΔH = Eₐ(forward) − Eₐ(reverse)
Why a catalyst does not shift equilibrium position or change K_eq

Understand

Why decreasing temperature shifts an exothermic reaction to the right using E_a reasoning
Why the NO₂/N₂O₄ system changes colour with temperature and pressure
How to draw and interpret rate-vs-time graphs for systems after disturbances

Can Do

Apply collision theory language to explain equilibrium approach and disturbances
Draw energy diagrams with correct E_a(forward) and E_a(reverse) for exo- and endothermic reactions
Explain why a catalyst is industrially valuable even though it doesn't change yield

Key Terms, scan these before reading

Collision theory
States that reactions occur when particles collide with sufficient energy and correct orientation.

Activation energy (Ea)
The minimum energy required for a collision to result in a chemical reaction.

Maxwell-Boltzmann distribution
A graph showing the spread of particle kinetic energies in a gas or solution at a given temperature.

Effective collision
A collision that results in bond breaking and formation, producing products.

Frequency factor
The number of collisions per unit time between reactant particles.

Temperature effect on rate
Increasing temperature increases the proportion of molecules with energy ≥ Ea, raising reaction rate.

Misconceptions to Fix

✗ Wrong: A catalyst increases the amount of product formed at equilibrium.

✓ Right: A catalyst speeds up both forward and reverse reactions equally, so equilibrium is reached faster but the equilibrium position and product yield do not change. Catalysts lower activation energy but do not alter ΔH or the thermodynamic favourability of a reaction.

Predict then reveal+8 XP

1 · Predict

2 · Reveal

3 · Compare

At the start of a reversible reaction (only reactants present), the forward rate is high and the reverse rate is zero. Predict: as the reaction proceeds toward equilibrium, what happens to these two rates, and why?

Confidence 50%

Collision theory revisited

Core Content

Collision Theory Revisited, From Module 3 to Module 5

Collision theory is the particle-level language for everything in equilibrium chemistry, it explains not just whether reactions happen, but how fast, and why the rates change as the system evolves.

Interactive, Equilibrium Simulation

Use the simulator. Let it run from 20 A and 0 B until the equilibrium badge appears, then read the rate display. Which statement matches what you see?

From Module 3, you learned that for a reaction to occur, particles must collide with correct orientation and sufficient energy (≥ the activation energy E_a). The rate of reaction depends on the frequency of effective collisions.

In Module 5, this same framework explains the approach to dynamic equilibrium. As a reversible reaction proceeds forward, three things change simultaneously:

The concentration of reactants decreases → fewer reactant-reactant collisions per second → forward rate decreases
The concentration of products increases → more product-product collisions per second → reverse rate increases
The rates converge until forward rate = reverse rate, which is dynamic equilibrium

This is not coincidence, it is a direct consequence of collision theory applied to a reversible system.

HSC language

In HSC extended response answers about equilibrium approach, use collision theory explicitly: "As reactants are consumed, the concentration of reactants decreases, reducing the frequency of effective collisions in the forward direction. Simultaneously, as products accumulate, the frequency of effective reverse collisions increases. When these rates equalise, dynamic equilibrium is established." This level of detail is required for Band 5–6.

Common error

Students say "the reaction slows down as equilibrium is approached." This is imprecise and loses marks. The FORWARD reaction slows down, but the REVERSE reaction speeds up. At equilibrium, neither has stopped, both are occurring at equal, non-zero rates. Always distinguish between forward and reverse rates.

Sequence showing forward/reverse rates approaching equilibrium

The four stages of approaching dynamic equilibrium, collision theory language at each stage

What to write in your book

Forward rate ∝ [reactants], starts at maximum, decreases as reactants consumed
Reverse rate ∝ [products], starts at zero, increases as products accumulate
Dynamic equilibrium: forward rate = reverse rate ≠ 0
HSC language: always say "frequency of effective collisions decreases/increases"

When a reversible reaction starts with only pure reactants, which of the following correctly describes the initial state?

Activation energy and equilibrium position

Activation Energy and the Equilibrium Position

The activation energy diagrams you studied in Module 4 take on new meaning in equilibrium chemistry, the relative heights of the forward and reverse energy barriers determine which direction is naturally preferred.

An energy diagram for a reversible reaction shows two activation energies:

E_a(forward) energy barrier from reactants to the transition state
E_a(reverse) energy barrier from products to the transition state

The difference is the enthalpy change: $\Delta H = E_a(\text{forward}) - E_a(\text{reverse})$

For an exothermic forward reaction (ΔH < 0): Products are lower in energy than reactants. The energy barrier going forward (E_a forward) is smaller than the barrier going backward (E_a reverse). At a given temperature, a larger fraction of particles can cross the forward barrier → forward rate is inherently faster at the same concentrations → equilibrium position lies on the products side.

For an endothermic forward reaction (ΔH > 0): Products are higher in energy than reactants. E_a(reverse) is smaller than E_a(forward) → more particles can cross the reverse barrier → equilibrium position lies on the reactants side.

Energy profile for an exothermic forward reaction, products are lower in energy, so E_a(forward) < E_a(reverse)

Must know

The relationship ΔH = E_a(forward) − E_a(reverse) is important. For an exothermic reaction, ΔH is negative because E_a(forward) < E_a(reverse). Make sure you can draw and label this energy diagram correctly for both exothermic and endothermic cases.

Common error

Students draw both activation energies the same height for exothermic reactions, assuming "exothermic just means energy is released." For an exothermic reaction, E_a(forward) < E_a(reverse). This difference is exactly equal to |ΔH|.

What to write in your book

ΔH = Eₐ(forward) − Eₐ(reverse)
Exothermic forward: Eₐ(fwd) < Eₐ(rev), products sit lower in energy
Endothermic forward: Eₐ(fwd) > Eₐ(rev), products sit higher in energy
Cooling shifts exothermic equilibrium RIGHT (reverse rate falls proportionally more)

For an exothermic forward reaction, E_a(forward) is greater than E_a(reverse).

Why catalysts do not shift equilibrium

Why Catalysts Don't Shift Equilibrium Position

A catalyst is the great equaliser of equilibrium chemistry, it accelerates both directions of a reversible reaction proportionally, getting you to equilibrium faster but leaving the destination unchanged.

A catalyst works by providing an alternative reaction pathway with a lower activation energy. In a reversible reaction, a catalyst lowers the E_a for BOTH the forward and reverse reactions by the same amount. Because both barriers are lowered equally:

The rate of the forward reaction increases by the same factor as the rate of the reverse reaction
The ratio of forward to reverse rates is unchanged and this ratio determines K_eq
Therefore: NO change to equilibrium position; NO change to K_eq
Equilibrium is reached more quickly (both rates are faster), but the destination is the same

HSC exam answer

"A catalyst is added to a reaction at equilibrium. Describe the effect on equilibrium position and K_eq." Correct answer: no effect on either. The catalyst increases the rate of both forward and reverse reactions equally, so if the system is already at equilibrium, no macroscopic shift occurs. K_eq depends only on temperature.

Most common error in Module 5

"Adding a catalyst shifts the equilibrium to the right." This is completely wrong. Catalysts affect the rate of reaching equilibrium, not the position of equilibrium. If the system is already at equilibrium, adding a catalyst changes nothing macroscopically.

Insight, Haber process

This is why the iron catalyst in the Haber process does not improve the yield of ammonia, only pressure and temperature can do that. The catalyst only allows the equilibrium position to be reached more quickly at the chosen temperature and pressure. Industrial chemists use catalysts to make the process economically viable (fast rate), not to improve the thermodynamic yield.

What to write in your book

Catalyst lowers Eₐ equally for BOTH forward AND reverse reactions
NO change to equilibrium position; NO change to K_eq
Increases RATE of reaching equilibrium only
K_eq depends only on temperature, a catalyst cannot change K_eq

Complete: A catalyst lowers Eₐ equally for both forward and reverse reactions, so the equilibrium ___ is unchanged even though equilibrium is reached faster.

The NO₂ and N₂O₄ equilibrium

The NO₂/N₂O₄ Equilibrium, Collision Theory in Action

The equilibrium between nitrogen dioxide and dinitrogen tetroxide is chemistry's most photogenic demonstration of collision theory at work, you can watch equilibrium shift in real time just by changing the temperature or pressure.

The equilibrium: 2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ/mol (exothermic)

NO₂ is brown/reddish-brown (dark colour indicates more NO₂)
N₂O₄ is colourless (paler mixture indicates more N₂O₄)

Temperature effect (collision theory): The forward reaction is exothermic → E_a(forward) < E_a(reverse). At lower temperature, fewer particles exceed either activation energy, but proportionally more can still cross the lower forward barrier. The forward rate decreases less than the reverse rate → forward rate > reverse rate → equilibrium shifts right → more N₂O₄ forms → mixture becomes paler when cooled.

Pressure effect: Increasing pressure by compressing the gas increases the concentration of all species → both forward and reverse collision frequencies increase. But the forward reaction converts 2 moles of NO₂ into 1 mole of N₂O₄, reducing total gas molecules. The forward rate increases more → equilibrium shifts toward fewer moles of gas (right) → more N₂O₄ → paler at higher pressure.

Must know values

For the NO₂/N₂O₄ system: NO₂ is brown; N₂O₄ is colourless; forward reaction (2NO₂ → N₂O₄) is exothermic (ΔH = −57 kJ/mol); 2 moles of gas on left, 1 mole on right. These facts appear repeatedly across Module 5.

Common error

Students describe cooling the NO₂/N₂O₄ system as "slowing the reaction down." Cooling slows BOTH reactions but slows the endothermic reverse reaction proportionally more than the exothermic forward reaction, so the forward rate becomes greater than the reverse, and the mixture shifts to form more N₂O₄ (paler). The direction of shift is determined by the relative activation energies.

What to write in your book

2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ/mol (exothermic forward)
NO₂ = brown; N₂O₄ = colourless
Cool → paler (forward rate decreases less than reverse → shift right → more N₂O₄)
Compress → paler (shift toward fewer gas moles = right)

A sealed syringe containing a NO₂/N₂O₄ equilibrium mixture is placed in hot water. What happens to the colour, and why?

Cross-lesson links: Le Chatelier's observation about the NO₂/N₂O₄ colour change introduced here is the basis of L05 (Le Chatelier's Principle, concentration and temperature). The activation energy argument explaining why temperature shifts exothermic equilibria left recurs in L07 (industrial applications). Rate-vs-time graphs in Card 5 are a core L01 and L06 skill tested with new disturbances.

Rate-vs-time graphs

Rate-vs-Time Graphs, Reading Equilibrium Approach

A rate-vs-time graph tells you the entire story of how a system approaches equilibrium, and knowing how to read and draw one is a core skill that distinguishes Band 4 answers from Band 6 answers in Module 5.

Three types of scenarios you must be able to draw and interpret:

Scenario	What the graph shows	Why
Approach from pure reactants	Forward rate starts high, decreases; reverse rate starts at zero, increases; both meet at non-zero plateau	Forward collisions decrease as reactants consumed; reverse collisions increase as products form
Approach from pure products	Mirror image: reverse rate starts high, decreases; forward rate starts at zero, increases; both meet at same plateau	Reverse collisions decrease as products consumed; forward collisions increase as reactants form
Adding reactant to a system at equilibrium	Forward rate spikes up suddenly; reverse rate unchanged initially; both re-equalise at a new higher plateau	More reactant → more forward collisions immediately; reverse catches up as more product forms

Key: the same equilibrium rate plateau is reached regardless of direction of approach (as long as total composition is the same). The equilibrium rate value represents the equal forward and reverse rates, both are non-zero.

Must know

Three types of rate-vs-time graph scenarios to master: (1) approach from pure reactants; (2) approach from pure products; (3) system at equilibrium with a disturbance applied (concentration, temperature, or pressure change). Practice all three.

Insight

The rate-vs-time graph for a system approaching equilibrium from pure PRODUCTS looks like the mirror image of the approach from pure reactants: reverse rate starts high and decreases; forward rate starts at zero and increases; they meet at the same equilibrium rate value. The same equilibrium position is reached from both directions.

What to write in your book

Rate-vs-time: from reactants, forward rate starts max, falls; reverse rate starts 0, rises; they meet
From products: mirror image, reverse rate starts max, forward starts 0; same plateau reached
Disturbance: adding reactant spikes forward rate; system re-equalises at higher plateau
Same equilibrium plateau regardless of approach direction

The same equilibrium rate plateau is reached regardless of whether the system starts from pure reactants or pure products (given the same total atomic composition).

✓

Worked examples and activities

✏️ Worked Examples

Worked Example 1, Applying collision theory to explain equilibrium approach

The reaction PCl₅(g) ⇌ PCl₃(g) + Cl₂(g) is endothermic (ΔH = +88 kJ/mol). A sealed flask is filled with pure PCl₅(g).

(a) Explain, using collision theory, how the system approaches dynamic equilibrium from this initial condition.
(b) Indicate which activation energy is larger, E_a(forward) or E_a(reverse), and explain why using the sign of ΔH.

At t = 0, only PCl₅ is present. Concentration of PCl₅ is at maximum → frequency of effective forward collisions is at maximum → forward rate is at its maximum. There are no products (PCl₃ and Cl₂) → reverse rate is zero.

As the reaction proceeds: PCl₅ is consumed → its concentration decreases → frequency of effective forward collisions decreases → forward rate decreases. PCl₃ and Cl₂ are produced → their concentrations increase → frequency of effective reverse collisions increases → reverse rate increases.

When forward rate = reverse rate (both non-zero), dynamic equilibrium is established. Concentrations of all species become constant.

The forward reaction is endothermic (ΔH = +88 kJ/mol) → products are higher in energy than reactants → the energy barrier going forward (from low-energy reactants up to the transition state) is larger than the barrier going backward (from high-energy products up to the transition state).

Using ΔH = E_a(forward) − E_a(reverse) = +88 kJ/mol → confirms E_a(forward) > E_a(reverse).

→ E_a(forward) > E_a(reverse)

Summary: (a) Forward rate starts maximum and decreases as PCl₅ is consumed; reverse rate starts at zero and increases as PCl₃ and Cl₂ accumulate; equilibrium established when rates equalise. (b) E_a(forward) > E_a(reverse) because the endothermic forward reaction means products sit higher in energy, it takes more energy to climb the hill going forward than backward.

Worked Example 2, Catalyst, temperature, and pressure effects in the Contact process

The Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ/mol. The system is at equilibrium.

(a) Using a reaction energy diagram, explain why the V₂O₅ catalyst does not shift the equilibrium position.
(b) Explain why the catalyst is still valuable industrially.
(c) Decreasing temperature shifts equilibrium to the right for this exothermic reaction. Explain using E_a and collision theory.

The V₂O₅ catalyst provides an alternative lower-energy pathway. On the energy diagram, the transition state peak is lowered by the same amount for both forward and reverse reactions. Both E_a(forward) and E_a(reverse) decrease by the same value x.

Because both barriers are lowered equally, the rate of the forward reaction increases by the same factor as the rate of the reverse reaction. The ratio of forward to reverse rates is unchanged, and it is this ratio that determines K_eq and equilibrium position.

→ No shift in equilibrium position; K_eq unchanged.

Although the catalyst does not improve yield, it dramatically increases the rate at which equilibrium is reached. At 450°C, the uncatalysed reaction would be too slow to be economically viable. The catalyst allows equilibrium to be reached quickly at a moderate temperature, making the process commercially feasible.

Without the catalyst, a higher temperature would be needed for sufficient rate, but this would decrease yield (exothermic reaction shifts left at higher T). The catalyst allows the compromise of moderate temperature with acceptable rate.

The forward reaction (SO₂ + O₂ → SO₃) is exothermic → E_a(forward) < E_a(reverse). At lower temperature, average kinetic energy of particles decreases, both forward and reverse rates decrease.

However, the rate decreases proportionally more for the reaction with the higher E_a. Since E_a(reverse) > E_a(forward), the reverse rate decreases more than the forward rate. Now forward rate > reverse rate → net forward reaction → equilibrium shifts right → more SO₃ produced.

Summary: (a) Catalyst lowers E_a equally for both directions, ratio of rates unchanged, no shift in equilibrium position or K_eq. (b) Catalyst enables acceptable rate at lower temperature, preserving higher yield from the exothermic equilibrium. (c) Decreasing T reduces both rates but reduces the reverse rate more (higher E_a reverse for exothermic reaction) → forward rate temporarily exceeds reverse → shift right to more SO₃.

Copy Into Your Books, Full Summary

Collision Theory and Equilibrium

Forward rate ∝ [reactants], decreases as reactants consumed
Reverse rate ∝ [products], increases as products accumulate
Equilibrium: forward rate = reverse rate (both non-zero)
HSC language: "frequency of effective collisions decreases/increases"

Activation Energy Rules

ΔH = Eₐ(forward) − Eₐ(reverse)
Exothermic forward: Eₐ(fwd) < Eₐ(rev), products lower energy
Endothermic forward: Eₐ(fwd) > Eₐ(rev), products higher energy
Cooling shifts exothermic equilibrium right (reverse rate falls more)

Catalyst Rules, Critical

Lowers Eₐ equally for BOTH forward AND reverse reactions
NO change to equilibrium position
NO change to K_eq
Increases RATE of reaching equilibrium only

NO₂/N₂O₄ System

2NO₂(g) ⇌ N₂O₄(g), ΔH = −57 kJ/mol
NO₂ = brown; N₂O₄ = colourless
Cool → paler (shift right, more N₂O₄)
Compress → paler (shift toward fewer gas moles = right)

Activities

Energy Diagrams for Equilibrium Reactions

For each reaction, describe a correctly labelled energy profile diagram including the relative positions of reactants, products, transition state, Eₐ(forward), Eₐ(reverse), and ΔH arrow.

2SO₂(g) + O₂(g) ⇌ 2SO₃(g), ΔH = −196 kJ/mol (exothermic forward). State which Eₐ is larger and why.
N₂(g) + O₂(g) ⇌ 2NO(g), ΔH = +180 kJ/mol (endothermic forward). State which Eₐ is larger and why.
Add a catalyst to the diagram for reaction 1. Describe what changes and what stays the same. Explain the effect on equilibrium position.

✓

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

✍️ Short Answer

UnderstandRecall(3 marks) Q6. Explain, using collision theory, why a catalyst does not change the position of equilibrium or the value of K_eq for a reversible reaction.

ApplyApply(4 marks) Q7. The equilibrium 2NO₂(g) ⇌ N₂O₄(g) is established in a sealed container at 25°C. The container is placed in a hot water bath at 80°C.
(a) Using collision theory and activation energy, explain why the mixture becomes darker brown at 80°C. (3 marks)
(b) State whether K_eq increases, decreases, or stays the same when temperature increases. Justify. (1 mark)

EvaluateAnalyse(7 marks) Q8. Extended Response: A chemist studies the equilibrium CO(g) + 3H₂(g) ⇌ CH₄(g) + H₂O(g), ΔH = −206 kJ/mol in a sealed vessel. The system is at equilibrium. Describe and explain, using collision theory and the concept of activation energy, the effect on the equilibrium position and on K_eq when: (a) the temperature is increased; (b) a nickel catalyst is added; (c) the volume of the vessel is halved.

Show all answers

MC Explanations

Approach to equilibrium from pure reactants. Forward rate starts at maximum and decreases (reactants consumed, fewer collisions). Reverse rate starts at zero and increases (products accumulate, more reverse collisions). Both converge to a non-zero equilibrium rate. The distractors are wrong because the forward rate never rises from a low start, and neither rate falls to zero at equilibrium.

Relative activation energies for an exothermic reaction. Exothermic: products are at lower energy → transition state is closer in energy to products → E_a(reverse) is larger. ΔH = E_a(fwd) − E_a(rev) = negative → E_a(fwd) < E_a(rev). Saying the two barriers are equal is wrong: the transition state is the same point, but the energy gaps up to it from the reactant side and the product side differ.

Effect of the V₂O₅ catalyst. Catalyst lowers E_a equally for both directions; both rates increase by the same factor; since rates were already equal, they remain equal; no shift. K_eq unchanged, it depends only on temperature. Every distractor incorrectly claims the equilibrium position or K_eq changes.

Compressing the NO₂/N₂O₄ syringe. Immediately after compression, [NO₂] doubles → mixture is momentarily darker. But the forward reaction (2 → 1 mol gas) has its collision frequency increased more than the reverse → forward rate > reverse rate → equilibrium shifts right → [NO₂] decreases, [N₂O₄] increases → mixture becomes paler than original.

Evaluating the Haber temperature-and-catalyst claim. Temperature: correctly increases rate (faster collisions), but DECREASES yield for this exothermic reaction (shift left at higher T). Catalyst: does NOT change yield, only increases the rate of reaching equilibrium without shifting the equilibrium position.

Short Answer Model Answers

Q6 (3 marks): A catalyst provides an alternative lower-energy reaction pathway, it lowers the activation energy [1]. For a reversible reaction, the catalyst lowers E_a for both the forward AND reverse reactions by the same amount [1]. Therefore, both the forward and reverse reaction rates increase by the same factor. Since the ratio of rates is unchanged, the equilibrium position is unchanged and K_eq is unchanged, K_eq depends only on temperature [1].

Q7 (4 marks): (a) 2NO₂ → N₂O₄ is exothermic (ΔH = −57 kJ/mol) → E_a(forward) < E_a(reverse) [1]. Increasing temperature increases the average kinetic energy of all particles, both forward and reverse reaction rates increase [1]. However, the rate increases proportionally more for the reaction with the higher activation energy. Since E_a(reverse) > E_a(forward), the reverse rate (N₂O₄ → 2NO₂) increases more than the forward rate, reverse rate > forward rate, net reverse reaction, equilibrium shifts left, more NO₂ is produced, mixture becomes darker brown [1]. (b) K_eq decreases, for an exothermic forward reaction, increasing temperature favours the reverse (endothermic) reaction, shifting equilibrium left and reducing the ratio of products to reactants, K_eq = [products]/[reactants] decreases [1].

Q8 (7 marks): (a) ΔH = −206 kJ/mol → exothermic forward → E_a(fwd) < E_a(rev) [1]. Increasing T → both rates increase → but reverse rate increases more (higher E_a) [1] → reverse > forward → shift LEFT → less CH₄ and H₂O, more CO and H₂ [1]. K_eq decreases [½]. (b) Ni catalyst lowers E_a equally for both forward and reverse directions [1]. Both rates increase by the same factor, system already at equilibrium remains at equilibrium; no shift; equilibrium position unchanged; K_eq unchanged [1]. (c) Halving volume doubles all concentrations → both rates increase → forward reaction involves 4 mol gas (CO + 3H₂); reverse involves 2 mol gas (CH₄ + H₂O) → forward collision frequency increases more [1] → forward > reverse → shift RIGHT → more CH₄ and H₂O; K_eq unchanged, pressure does not change K_eq [½].

✓

Retrieve and reflect

Check what actually stuck

Extended Response

The industrial chemist's dilemma: "I need to maximise the yield of SO₃ from the Contact process (2SO₂ + O₂ ⇌ 2SO₃, ΔH = −196 kJ/mol), but I also need the reaction to proceed at an acceptable rate." Using collision theory, activation energy concepts, and the role of catalysts, write a response explaining the rate versus yield trade-off and how the V₂O₅ catalyst resolves the industrial problem. (6 marks)

How did your thinking change?

Go back to your Think First prediction. Recall that in 1884, Le Chatelier at the École des Mines recorded exactly this result, he cooled a sealed N₂O₄/NO₂ tube and the brown colour faded. Now you have the collision theory explanation he needed.
• Was your prediction correct (paler)? Can you now give the full collision theory and E_a explanation of why Le Chatelier's observation was correct?
• Can you explain why the same temperature decrease would have the opposite colour effect if the forward reaction were endothermic instead?

I have completed this lesson

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