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Module 4 · L5 of 13 ~35 min ⚡ +50 XP in Learn · +25 to complete

Activation Energy, Catalysts & Energy Diagrams

Modern cars carry a catalytic converter, a precious-metal coating (platinum, palladium and rhodium) on a ceramic honeycomb. At exhaust temperatures the catalyst surface sharply lowers the activation energy for oxidising carbon monoxide, so a reaction that barely runs in open air becomes fast and efficient. The ΔH of combustion is unchanged; only E_a is lowered.

Today's hook, A catalytic converter coats a ceramic honeycomb with platinum and palladium. Its surface sharply lowers the activation energy for converting toxic exhaust gases, turning a reaction that barely runs into a fast one, while ΔH is completely unchanged. How does lowering one number on an energy diagram transform an emission problem?

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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

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

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Where this lesson fits

Syllabus requirement

Lesson question: how can lowering just one number on an energy diagram, the activation energy (E_a), turn a reaction that barely runs into a fast one, without changing how much energy the reaction releases?

1Place E_a. Activation energy is the climb from the reactant level up to the peak (the transition state) on an energy profile diagram.
2Add a catalyst. It offers an alternative pathway with a lower peak, so a greater fraction of collisions have enough energy to react.
3Check what stays fixed. Reactants, products and ΔH do not move; only E_a (and therefore the rate) changes.

Quick prerequisite: activation energy is the minimum energy for a successful collision (Module 3 collision theory), and an energy profile diagram plots enthalpy against reaction progress (Lesson 1). If either is shaky, start with the Supported route later in this lesson. Syllabus reference: construct energy profile diagrams to represent and analyse enthalpy changes and activation energy, and model the role of catalysts (ACSCH072, ACSCH073).

Know what matters most

Must know

A catalyst lowers E_a by providing an alternative pathway; it does NOT change ΔH, reactants or products
E_a is measured from the reactant level to the peak; a catalyst lowers the peak only
Catalysed and uncatalysed pathways share identical reactant and product levels and the same ΔH

Should know

E_a(reverse) = E_a(forward) − ΔH
Homogeneous (same phase as reactants) vs heterogeneous (different phase) catalysts
A catalyst speeds forward and reverse equally, so it reaches equilibrium faster but does not shift it or raise yield

Going deeper

Heterogeneous catalysts work by adsorbing reactants onto their surface
Enzymes as highly specific biological catalysts (active-site geometry)

Warm up and recall

Recall, your gut answer first

+5 XP warm-up

A catalytic converter on a modern car contains a thin honeycomb of platinum and palladium. Every day, highly toxic gases (carbon monoxide, unburnt hydrocarbons, nitrogen oxides) flow through it at high temperature. They are converted to carbon dioxide, water, and nitrogen, far less harmful products. After 200,000 km, the platinum is still there, completely intact.

Two common claims about catalysts:

A catalyst makes a reaction release more energy, it improves the energetics of the reaction.
A catalyst lowers the amount of energy the reactants need to possess before they can successfully react.

Which claim is correct? Which is false? Commit to an answer now, before any theory is shown.

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What you'll master, and the words for it

What you'll master

Know

Key Facts

E_a = minimum energy for a collision to result in reaction
A catalyst lowers E_a — it does NOT change ΔH, reactants, or products
Homogeneous catalyst: same phase as reactants; heterogeneous: different phase

Understand

Concepts

Why E_a and ΔH are independent quantities (kinetics vs thermodynamics)
How a catalyst provides an alternative mechanism with lower energy steps
Why a catalyst lowers E_a of the reverse reaction equally

Can Do

Skills

Draw and interpret catalysed vs uncatalysed energy profile diagrams
Calculate E_a(reverse) from E_a(forward) and ΔH
Classify a catalyst as homogeneous or heterogeneous with justification

Key terms

Activation energy (Ea)

The minimum energy required for reactants to form the transition state and proceed to products.

Transition state (activated complex)

The high-energy, unstable species at the peak of the energy profile; exists only briefly.

Catalyst

A substance that provides an alternative reaction pathway with lower activation energy; increases rate without altering ΔH or being consumed.

Catalysed vs uncatalysed Ea

A catalyst lowers Ea for both the forward and reverse reactions by the same amount; equilibrium position is unchanged.

Energy profile diagram

A graph of potential energy vs reaction progress; shows Ea (peak), ΔH (difference between reactant and product energy levels), and effect of catalysts.

Biological catalyst (enzyme)

A protein that acts as a highly specific catalyst in biochemical reactions; active site provides ideal geometry for transition state stabilisation.

Cross-lesson links: This lesson bridges L01 (energy profile diagrams showing ΔH) with L06–L10 (calculating ΔH). The key distinction is that a catalyst changes E_a but not ΔH, ΔH depends only on the start and end states (Hess's Law, L08), while E_a depends on the reaction mechanism. Biological catalysts (enzymes) return in L09, where the ATP coupling system uses Hess's Law to drive endothermic biosynthesis by pairing it with exothermic ATP hydrolysis.

How Catalysts Work, Lowering Ea

How Catalysts Work, Lowering E_a

Must know core concept

A catalyst provides an alternative reaction pathway with a lower activation energy, increasing the proportion of collisions that have enough energy to react, without being consumed in the process.

A catalyst works by providing an alternative mechanism, a different sequence of bond-breaking and bond-forming steps, each of which requires less energy at each stage. On an energy profile diagram, the catalysed pathway shows a lower peak (lower E_a) but the same starting point (reactants) and the same ending point (products).

Both pathways begin at the same reactant level and end at the same product level, ΔH is identical. The catalyst lowers E_a by providing an alternative mechanism.

What a catalyst does and does not change:

Property	With catalyst	Without catalyst
E_a (forward)	Lower	Higher
E_a (reverse)	Lower (same reduction)	Higher
ΔH	Unchanged	Same
Reactants	Unchanged	Same
Products	Unchanged	Same
Reaction rate	Faster	Slower
Catalyst consumed?	No, regenerated	Not applicable

A catalyst lowers E_a only. It does NOT change ΔH, the identity of reactants, or the identity of products. If any question or statement says a catalyst changes ΔH, that statement is false.

Common error: "The catalyst provides energy for the reaction.", Incorrect. The catalyst lowers the energy barrier; it does not supply energy. Reactant molecules still need to reach the (lower) transition state through thermal collisions. The catalyst just makes that threshold more achievable.

Two-column comparison of what a catalyst changes versus what it leaves unchanged in a reaction.

A catalyst provides an alternative reaction pathway with a lower activation energy (E_a), increasing the proportion of successful collisions; the catalyst takes part in the mechanism and is regenerated overall, so it is not used up. On an energy profile diagram the catalysed pathway shows a lower peak; ΔH is unchanged.

Common misconception, catalysts and yield. A catalyst lowers E_a for the forward and reverse reactions by the same amount, so it speeds both equally. It lets a reaction reach equilibrium faster but does not shift the equilibrium position or increase the maximum yield, and it never changes ΔH.

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

Quick check: A catalyst is added to an exothermic reaction. Which of the following is correct?

Homogeneous vs Heterogeneous Catalysts

Should know core concept

We just saw that catalysts lower E_a without changing ΔH. That raises a question: are all catalysts the same, or do they work by fundamentally different mechanisms? This card answers it → by distinguishing homogeneous (same phase) from heterogeneous (different phase) catalysts.

Catalysts are classified by whether they are in the same physical state (phase) as the reactants, homogeneous, or a different phase, heterogeneous.

Compares homogeneous catalysts (same phase as reactants) with heterogeneous catalysts (different phase) using esterification and catalytic-converter examples. — Left: homogeneous, H⁺(aq) catalyst in the same aqueous phase as the reactants. Right: heterogeneous, solid platinum catalyses gaseous exhaust molecules; reactions occur at the solid surface.

Catalyst	Type	Reaction	Phase of catalyst → reactants
Pt, Pd, Rh	Heterogeneous	Catalytic converter: CO + hydrocarbons + NO_x → CO₂, N₂, H₂O	Solid → gas
Fe	Heterogeneous	Haber process: N₂(g) + 3H₂(g) → 2NH₃(g)	Solid → gas
H⁺(aq)	Homogeneous	Esterification of CH₃COOH(aq) + C₂H₅OH(aq)	Aqueous → aqueous
MnO₄⁻(aq)	Homogeneous	Oxidation reactions in aqueous solution	Aqueous → aqueous

When asked to classify a catalyst: state the physical state of the catalyst AND the physical state of the reactants, then justify. Saying "heterogeneous" without stating that the phases differ loses marks in HSC.

Common error: Confusing a catalyst with a reactant. A catalyst participates in the mechanism and is regenerated, it does not appear in the overall balanced equation as a consumed reagent.

Comparison table of homogeneous vs heterogeneous catalysts across five properties.

Homogeneous catalysts are in the same phase as the reactants (e.g. H⁺(aq) in an aqueous reaction); heterogeneous catalysts are in a different phase (e.g. solid Pt catalysing a gas-phase reaction). Heterogeneous catalysts work by adsorbing reactants onto their surface.

Add the highlighted point to your notes before the check below.

Explain it: Vanadium(V) oxide (V₂O₅) is used as a catalyst in the Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). V₂O₅ is a solid. In two sentences, classify V₂O₅ as homogeneous or heterogeneous and justify your answer.

Drawing Catalysed vs Uncatalysed Diagrams

Must know core concept

We just saw that homo- and heterogeneous catalysts both lower E_a by different mechanisms. That raises a question: how do we show both the catalysed and uncatalysed pathways on a single energy profile diagram? This card answers it → with two peaks at different heights sharing the same reactant and product levels.

A single diagram can show both pathways, two peaks at different heights, but the same reactant level, the same product level, and the same ΔH arrow.

Rules for drawing a correct catalysed comparison:

Draw both curves starting from the same reactant energy level
Draw both curves ending at the same product energy level
Uncatalysed curve: taller peak (higher E_a)
Catalysed curve: lower peak (lower E_a), sometimes shows multiple smaller humps (multi-step mechanism)
Label E_a(uncatalysed) and E_a(catalysed) from the reactant level to each respective peak
Draw the ΔH arrow from the reactant level to the product level, identical for both pathways

ΔH arrow must be identical in both pathways. Draw it from the reactant level to the product level, not to the peak. If your ΔH arrow changes between catalysed and uncatalysed diagrams, the diagram is wrong.

Critical error to avoid: Drawing the product level lower for the catalysed pathway. This implies ΔH changes with a catalyst, it does not. Products are identical; their energy level is identical. The only thing that changes is the height of the peak (E_a).

CATALYSED vs UNCATALYSED ENERGY PROFILE, INTERACTIVE Interactive

Use Catalyst: ON to show both pathways. Observe: same ΔH, lower E_a for catalysed route.

Going deeper: how a catalytic converter applies this

Real-World Anchor, Catalytic Converters: The catalytic converter in a modern car contains a ceramic honeycomb coated with platinum, palladium, and rhodium. Exhaust gases (CO, unburnt hydrocarbons, NO_x) pass over the solid metal surface at high temperature.

The solid catalyst provides a surface for the gases to adsorb onto, lowering the E_a for reactions that convert these toxic species to CO₂, H₂O, and N₂. After long service, the platinum is still there because it is regenerated after each cycle of adsorption and reaction, even though it takes part in the mechanism. ΔH for each reaction is unchanged by the catalyst, the converter does not affect how much energy the exhaust reactions release.

A catalysed energy profile diagram shows two pathways: uncatalysed (higher peak, larger E_a) and catalysed (lower peak, smaller E_a). Both share identical reactant and product levels and the same ΔH, only activation energy differs.

Pause, write the highlighted definition into your book.

Fill in the blanks: Complete the sentences about catalyst diagrams.

Both the catalysed and uncatalysed curves must start at the same ___ level and end at the same ___ level, because a catalyst does not change ___. The only difference is the height of the ___, which is lower for the ___ pathway.

Same science, choose your support

07b

Same science, choose your support

pick one

All three routes practise the same Must-Know idea: a catalyst lowers E_a without changing ΔH, and E_a is read from the reactant level to the peak. Do one; try another if you want more practice.

Supported word bank + sentence frame start here if unsure

Word bank: pathway · lower · peak · unchanged · reactant

An uncatalysed reaction has E_a = 150 kJ mol⁻¹ and ΔH = −60 kJ mol⁻¹. A catalyst is added. Complete the frame:

"The catalyst provides an alternative ______ with a ______ activation energy, so the ______ (transition state) drops. ΔH stays ______ because the ______ and product energy levels do not move."

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Show a hint

A catalyst only lowers the peak (E_a). The reactant and product levels are fixed, so ΔH = H(products) − H(reactants) cannot change.

Core independent HSC-standard the exam standard

A reaction has E_a(forward) = 150 kJ mol⁻¹ and ΔH = −60 kJ mol⁻¹. A catalyst lowers E_a by 55 kJ mol⁻¹. State the catalysed forward E_a, explain what happens to ΔH and why, and describe exactly where E_a is measured on the energy profile diagram.

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Stretch reasoning in a new context push yourself

Two students draw energy profile diagrams for the same catalysed reaction. One draws the catalysed products at a lower energy level than the uncatalysed products; the other keeps the E_a arrow starting from the product level. Explain why each diagram is wrong, using E_a and ΔH, and describe the single correct diagram.

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Worked examples

Worked examples · reveal as you go

Worked example 1 +5 XP on full reveal

Interpreting a catalysed energy profile diagram. A reaction has an uncatalysed E_a of 180 kJ mol⁻¹ and ΔH = −75 kJ mol⁻¹. In the presence of a catalyst, E_a drops to 95 kJ mol⁻¹.

(a) What is E_a of the reverse uncatalysed reaction?
(b) What is ΔH for the catalysed reaction?
(c) What is E_a of the catalysed reverse reaction?

GIVEN / FIND
GIVEN: E_a(forward, uncatalysed) = 180 kJ mol⁻¹ | ΔH = −75 kJ mol⁻¹ | E_a(forward, catalysed) = 95 kJ mol⁻¹
FIND: (a) E_a(reverse, uncat) | (b) ΔH(catalysed) | (c) E_a(reverse, catalysed)

Always identify all given information and what is being asked before calculating.

Step (a), E_a of reverse uncatalysed
E_a(reverse) = E_a(forward) − ΔH = 180 − (−75) = 180 + 75 = 255 kJ mol⁻¹

For the reverse reaction, you start from the products (which sit 75 kJ mol⁻¹ below reactants) and climb to the same peak. Since products are lower, the reverse reaction must climb further → E_a(reverse) is larger than E_a(forward) for an exothermic reaction.

Step (b), ΔH for catalysed reaction
ΔH(catalysed) = −75 kJ mol⁻¹ (unchanged)

A catalyst does not change ΔH. Reactants and products are identical, only the pathway changes. Same answer as the uncatalysed reaction.

Step (c), E_a of catalysed reverse reaction
E_a(reverse, cat) = E_a(forward, cat) − ΔH = 95 − (−75) = 95 + 75 = 170 kJ mol⁻¹

Final Answers: (a) 255 kJ mol⁻¹ | (b) −75 kJ mol⁻¹ (unchanged) | (c) 170 kJ mol⁻¹

Same logic applied to the catalysed diagram. Products still sit 75 kJ mol⁻¹ below reactants; the catalysed peak is now only 95 kJ mol⁻¹ above reactants, so the reverse reaction climbs 95 + 75 = 170 kJ mol⁻¹ from the product level.

Worked example 2 +5 XP on full reveal

Identifying homogeneous vs heterogeneous catalysts. The Haber process reacts N₂(g) and H₂(g) over an iron (Fe) catalyst to produce NH₃(g). In a second reaction, H⁺(aq) ions catalyse the reaction between CH₃COOH(aq) and C₂H₅OH(aq) to form an ester.

Identify whether each catalyst is homogeneous or heterogeneous, justify your answer, and state what effect each catalyst has on ΔH for its reaction.

GIVEN / FIND
GIVEN: Fe catalyst, N₂(g) + H₂(g) reactants | H⁺(aq) catalyst, CH₃COOH(aq) + C₂H₅OH(aq) reactants
FIND: Catalyst type for each; effect on ΔH

The key question is always: are the catalyst and reactants in the same physical state?

Step 1, Haber process catalyst
Fe is a solid; N₂ and H₂ are gases. Solid ≠ gas → different phases → heterogeneous catalyst.

The gas-phase reactants adsorb onto the surface of the solid iron catalyst, where their bonds weaken and the reaction proceeds with lower E_a.

Step 2, Esterification catalyst
H⁺ is an aqueous ion; both reactants are in aqueous solution. Aqueous = aqueous → same phase → homogeneous catalyst.

All species (catalyst and reactants) are dissolved in the same solvent. The H⁺ ion participates in the reaction mechanism and is regenerated.

Step 3, Effect on ΔH
Neither catalyst changes ΔH. For the Haber process, ΔH = −92 kJ mol⁻¹ with or without the Fe catalyst. For esterification, ΔH is unchanged by H⁺. Catalysts only lower E_a; they never alter the enthalpy of products or reactants.

E_a is a kinetic quantity (rate); ΔH is a thermodynamic quantity (energy). Catalysts affect kinetics only.

Key idea

Sort the steps +7 XP

Put the steps for reading activation energy from an energy profile diagram in the correct order.

Measure the energy difference from the reactants level to the peak (transition state).
Identify the reactants energy level on the y-axis.
Record this as E_a, always a positive value.
Identify the transition state at the highest point of the curve.
Note that ΔH is measured separately: products level minus reactants level.

What you'll master, and the words for it

Formula reference · this lesson

core formula

📐

Key Relationships, This Lesson

$E_a$

E_a = activation energy (kJ mol⁻¹), minimum energy required for reactant molecules to reach the transition state Measured from the reactant energy level to the peak of the energy profile diagram

$E_a(\text{cat}) < E_a(\text{uncat})$

A catalyst lowers E_a, provides a lower-energy alternative pathway ΔH is unchanged, reactants and products are identical; only the pathway changes

$E_a(\text{reverse}) = E_a(\text{forward}) - \Delta H$

For an exothermic reaction (ΔH negative): E_a(reverse) = E_a(forward) + |ΔH| → larger than forward E_a For an endothermic reaction (ΔH positive): E_a(reverse) = E_a(forward) − ΔH → smaller than forward E_a

Catalyst rule: Lowers E_a only | ΔH unchanged | Reactants unchanged | Products unchanged | Catalyst not consumed

Warm up and recall

02b

Printable worksheet

Use the PDF for classwork, homework or revision. It includes key ideas, activities, questions, an extend task and success-criteria proof.

Open printable version

"A catalyst changes ΔH, it makes the reaction release more energy"

Common errors · the 3 traps that cost marks

"A catalyst changes ΔH, it makes the reaction release more energy"

Students write "using a catalyst releases more energy per mole" or "ΔH becomes more negative with a catalyst." The logic seems intuitive: faster reaction = more energy.

Fix: A catalyst only lowers E_a. ΔH depends on the energy of reactants vs products, and since the catalyst leaves both unchanged, ΔH is identical with or without the catalyst. The reaction releases the same total energy; it just releases it faster.

"Ea(catalysed) is measured from the x-axis / product level"

Students draw the E_a arrow for the catalysed pathway starting from the x-axis (zero) or from the product level rather than from the reactant level.

Fix: E_a for the forward reaction is always measured from the reactant energy level to the transition state peak, regardless of whether the pathway is catalysed or uncatalysed. Measuring from the product level gives E_a of the reverse reaction, not the forward reaction.

"The catalysed pathway ends at a lower product energy level"

Students draw the catalysed curve ending at a lower product energy level, implying the catalyst changes where the products end up energetically.

Fix: Both curves must end at exactly the same product energy level. The products are chemically identical in both pathways, same species, same bonds, same enthalpy. Only the height of the transition state peak changes. If your diagram shows different product levels, it implies ΔH changes, which is wrong.

Drill, then revisit

Quick-fire practice · 5 reps +2 XP per reveal

A student drew an energy profile diagram with E_a(uncatalysed) drawn from the x-axis to the peak, and the catalysed product level lower than the uncatalysed product level. Identify both errors.

A reaction has E_a(forward, uncatalysed) = 210 kJ mol⁻¹ and ΔH = +55 kJ mol⁻¹. A catalyst reduces E_a by 80 kJ mol⁻¹. Calculate: (a) E_a(reverse, uncatalysed), (b) E_a(forward, catalysed), (c) E_a(reverse, catalysed).

Classify each catalyst as homogeneous or heterogeneous and justify: (a) MnO₄⁻(aq) in an aqueous oxidation reaction; (b) Pt(s) in the conversion of SO₂(g) to SO₃(g); (c) concentrated H₂SO₄(l) in the dehydration of C₂H₅OH(l).

A student argues: "Adding a catalyst to an endothermic reaction makes it exothermic, because the catalyst lowers E_a and the products are now at lower energy." Is this correct? Explain why or why not.

The uncatalysed oxidation of CO has E_a = 232 kJ mol⁻¹. The catalytic converter (Pt surface) reduces this to 75 kJ mol⁻¹. ΔH = −283 kJ mol⁻¹. Using the concept of E_a, explain why the converter is effective at exhaust temperatures (400–600°C) while the uncatalysed reaction is not.

Revisit your thinking

Go back to your Think First response. Now that you've studied catalysts and energy diagrams, revisit the 1975 EPA catalytic converter mandate, Ford and GM's platinum–palladium system dropping CO oxidation E_a from ~140 kJ mol⁻¹ to ~75 kJ mol⁻¹:

Claim 1, FALSE: "A catalyst makes a reaction release more energy." The 1975 catalytic converter does not change ΔH for CO oxidation. The total energy released per mole is identical, because the reactants (CO + ½O₂) and products (CO₂) are unchanged, and ΔH depends only on their relative energies.
Claim 2, CORRECT: "A catalyst lowers the amount of energy reactants need to possess before they can react." The platinum surface lowers E_a substantially, so at exhaust temperatures (400–600°C) a far greater proportion of CO molecules now have sufficient energy to react. The result is near-complete conversion rather than the very low conversion of the uncatalysed reaction.
Draw a labelled energy profile diagram for the catalysed and uncatalysed CO oxidation pathways. Mark E_a(uncatalysed) = 140 kJ mol⁻¹, E_a(catalysed) = 75 kJ mol⁻¹, and show that ΔH is identical for both.

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Interactive Tool, Enthalpy & Calorimetry Open fullscreen ↗

Use the Calorimetry Calculator. Heating 200 g of water by 5.0°C (c = 4.18 J/g·°C) requires how much heat?

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Multiple choice

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Pick your answer, then rate your confidence. That tells the system what to drill next.

01b

Misconceptions to fix before short answer

Wrong: A catalyst changes the enthalpy change (ΔH) of a reaction.

Right: A catalyst provides an alternative reaction pathway with lower activation energy but does not change the reactant or product energy levels. ΔH remains unchanged because it depends only on the difference between product and reactant enthalpies, not the pathway taken.

Wrong: E_a(catalysed) is measured from the x-axis or from the product level.

Right: E_a for the forward reaction is always measured from the reactant energy level to the transition state peak. Measuring from the product level gives E_a of the reverse reaction.

Wrong: The catalysed pathway ends at a lower product level because E_a is lower.

Right: Both pathways end at the same product level. The catalyst changes only the peak height, not the endpoint of the reaction.

Short answer

ApplyBand 4

Q6. A reaction has the following energy data: E_a(forward, uncatalysed) = 210 kJ mol⁻¹; ΔH = +55 kJ mol⁻¹. A catalyst reduces E_a by 80 kJ mol⁻¹.

(a) Is this reaction exothermic or endothermic? (1 mark)
(b) Calculate E_a(reverse, uncatalysed). (1 mark)
(c) Calculate E_a(forward, catalysed) and E_a(reverse, catalysed). (2 marks)
(d) What is ΔH for the catalysed reaction? Explain. (1 mark) 5 MARKS

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ApplyBand 4

Q7. The industrial production of sulfuric acid uses vanadium(V) oxide (V₂O₅) as a catalyst in the Contact process: 2SO₂(g) + O₂(g) ⇌ 2SO₃(g). V₂O₅ is a solid.

(a) Classify V₂O₅ as a homogeneous or heterogeneous catalyst. Justify your answer by referring to physical states. (2 marks)
(b) Explain, using the concept of activation energy, how V₂O₅ increases the rate of production of SO₃. (2 marks)
(c) A student suggests that using a catalyst will produce more SO₃ at equilibrium. Is this correct? Explain. (1 mark) 5 MARKS

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EvaluateBand 5

Q8. A catalytic converter in a car converts toxic exhaust gases to less harmful products. The converter contains platinum (Pt), palladium (Pd), and rhodium (Rh) embedded on a ceramic honeycomb surface.

(a) Identify the physical state of the catalyst and the physical state of the exhaust gases. Hence classify the catalytic converter as using a homogeneous or heterogeneous catalyst. (2 marks)
(b) The uncatalysed combustion of CO has E_a = 232 kJ mol⁻¹. The catalytic converter reduces this to 75 kJ mol⁻¹. Use the concept of activation energy to explain why the catalytic converter is effective at converting CO at typical exhaust temperatures (400–600°C), but a car without a converter rarely oxidises CO completely at these temperatures. (3 marks)
(c) Explain why the catalytic converter does not alter the total energy released by the combustion reactions in the exhaust. (1 mark) 6 MARKS

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Comprehensive Answers

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Short Answer Model Answers

Q6 (5 marks):
(a) ΔH = +55 kJ mol⁻¹ → positive → endothermic [1].
(b) E_a(reverse, uncat) = 210 − (+55) = 155 kJ mol⁻¹ [1]. (Products are 55 kJ mol⁻¹ higher than reactants; the reverse reaction starts from a higher level and needs less energy to reach the same peak.)
(c) E_a(forward, cat) = 210 − 80 = 130 kJ mol⁻¹ [1]; E_a(reverse, cat) = 130 − 55 = 75 kJ mol⁻¹ [1]. (The same reduction of 80 kJ mol⁻¹ applies to the peak; the reverse E_a is measured from the product level to the new lower peak.)
(d) ΔH(catalysed) = +55 kJ mol⁻¹, unchanged. A catalyst does not change ΔH because reactants and products are identical to the uncatalysed reaction; only the pathway (and hence E_a) is altered [1].

Q7 (5 marks):
(a) V₂O₅ is a solid; SO₂ and O₂ are gases [½]. Solid ≠ gas → different physical states → heterogeneous catalyst [1 + ½].
(b) V₂O₅ provides an alternative reaction mechanism with a lower activation energy than the uncatalysed pathway [1]. At operating temperature, a greater proportion of collisions now have combined energy at or above E_a(catalysed), so the frequency of successful collisions increases, increasing the rate of SO₃ production [1].
(c) No, the student is incorrect [½]. A catalyst changes the rate at which equilibrium is reached but does not change the position of equilibrium (the ratio of products to reactants at equilibrium). The equilibrium amount of SO₃ depends on ΔG and temperature, not on whether a catalyst is present [½].

Q8 (6 marks):
(a) Catalyst (Pt, Pd, Rh) = solid. Exhaust gases (CO, hydrocarbons, NO_x) = gas [½]. Solid ≠ gas → heterogeneous catalyst [1 + ½].
(b) Without converter: E_a = 232 kJ mol⁻¹. At 400–600°C, a relatively small proportion of collisions have combined energy at or above the 232 kJ mol⁻¹ activation barrier, so the uncatalysed oxidation rate is low [1]. With converter: the Pt surface provides an alternative mechanism (adsorption, surface reaction, desorption) with E_a = 75 kJ mol⁻¹ [1]. At the same exhaust temperature, a much greater proportion of molecules now have sufficient energy, so the rate of CO oxidation increases dramatically and conversion is essentially complete before the gas exits the converter [1].
(c) ΔH depends on the energies of the reactants and products, which are identical with or without the catalyst [½]. The catalyst only alters the pathway (E_a), not the starting or ending energy levels, so total energy released is unchanged [½].

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