Catalysis and Enzymes

Compare heterogeneous and homogeneous catalysts, and explain how enzymes depend on conditions.

GCE A-Level H2 Chemistry 9476-2027

On this page

Catalysis questions are usually “explain in one chain”: catalyst → lower Eₐ → larger fraction with E ≥ Eₐ → faster rate. Enzyme questions add biology-specific ideas (active site, optimum conditions, denaturation) but the kinetic logic is still the same.

Link this page to Rate Equations, Orders, and Rate Constant and the Reaction Kinetics hub to keep mechanism and data interpretation consistent.

Definitions (Must Know)

A. Catalyst

A catalyst increases reaction rate by providing an alternative pathway with a lower activation energy, Eₐ, and is regenerated (not used up overall).

B. Heterogeneous catalysis

Heterogeneous catalysis occurs when the catalyst is in a different phase from the reactants (often a solid catalyst with gaseous reactants). It typically involves adsorption onto the catalyst surface.

C. Homogeneous catalysis

Homogeneous catalysis occurs when the catalyst is in the same phase as the reactants. It typically involves intermediates in a catalytic cycle.

D. Adsorption

Adsorption is the sticking of reactant particles onto the surface of a solid catalyst at active sites.

E. Enzyme

An enzyme is a biological catalyst with a specific active site that binds substrates and catalyses a reaction.

F. Denaturation

Denaturation is a permanent change in the enzyme’s structure (active site shape), usually caused by high temperature or extreme pH, causing loss of catalytic activity.

Key Ideas (What Earns Marks)

A catalyst increases rate by lowering Eₐ (alternative pathway), not by “changing Δ H”.
A catalyst does not change equilibrium position; it speeds up both forward and reverse reactions.
Heterogeneous catalysts:
- have active sites on a surface
- often require adsorption → reaction → desorption
- higher surface area gives more active sites → faster rate
Homogeneous catalysts:
- form intermediates
- are regenerated at the end of the cycle
Enzymes:
- are highly specific (active site)
- have an optimum temperature and pH
- denature at high temperature / extreme pH

“A catalyst provides an alternative pathway with lower Eₐ, so a larger fraction of molecules have E ≥ Eₐ at the same temperature, giving more successful collisions per unit time.”

Detailed Explanations

A. Why catalysts don’t change equilibrium

A catalyst lowers the activation energy for both the forward and reverse reactions. Therefore equilibrium is reached faster, but the equilibrium position is unchanged.

B. Heterogeneous catalysis (surface mechanism)

Typical steps:

reactants adsorb onto the catalyst surface (at active sites)
bonds weaken / new bonds form on the surface (reaction occurs)
products desorb and leave the surface

Increasing surface area increases the number of active sites, so more reactant particles can react per unit time.

C. Homogeneous catalysis (intermediate cycle)

In homogeneous catalysis, the catalyst forms an intermediate and is regenerated.

Example (iodide-catalysed decomposition of hydrogen peroxide):

Step 1: H₂O₂(aq) + I⁻(aq) → H₂O(l) + IO⁻(aq)
Step 2: H₂O₂(aq) + IO⁻(aq) → H₂O(l) + O₂(g) + I⁻(aq)

Overall (add steps, cancel I⁻ and IO⁻): 2H₂O₂(aq) → 2H₂O(l) + O₂(g)

So I⁻ is the catalyst (regenerated) and IO⁻ is an intermediate.

D. Enzymes (optimum conditions and denaturation)

Enzymes are proteins that act as biological catalysts. In the lock-and-key model, only a substrate with a complementary shape can fit the enzyme’s active site. Binding forms an enzyme–substrate complex and provides a lower-activation-energy route. The products leave and the active site can be used again.

At low to moderate temperature, increasing temperature increases kinetic energy and collision frequency, so rate increases.

Above an enzyme’s optimum temperature, the enzyme denatures, changing the active site shape, so fewer enzyme–substrate complexes form and rate falls sharply.

Enzyme Activity vs Temperature (Typical Shape). Enzyme activity plotted as Reaction rate against Temperature. — Typical enzyme trend: rate increases with temperature up to an optimum, then drops sharply as the enzyme denatures (shape varies by enzyme).

Values for Enzyme Activity vs Temperature (Typical Shape)
Temperature (°C)	Enzyme activity
0	0.05
10	0.2
20	0.5
30	0.8
37	1
45	0.55
60	0.1
80	0

Values for Enzyme Activity vs Temperature (Typical Shape)
Temperature (°C)	Enzyme activity
0	0.05
10	0.2
20	0.5
30	0.8
37	1
45	0.55
60	0.1
80	0

E. Required catalytic examples

Use each example to connect the catalyst type to a lower-energy route:

Process	Catalyst and type	What to explain
Haber process, N₂ + 3H₂ ⇌ 2NH₃	iron, heterogeneous	nitrogen and hydrogen adsorb, their bonds weaken, ammonia forms and desorbs
Catalytic converter	platinum, palladium or rhodium surfaces, heterogeneous	harmful gases such as CO and NO adsorb and react to form less harmful products such as CO₂ and N₂
Oxidation of atmospheric SO₂	NO/NO₂ cycle, homogeneous	NO is oxidised to NO₂; NO₂ then oxidises SO₂ to SO₃ and regenerates NO
S₂O₈²⁻ with I⁻	Fe²⁺/Fe³⁺, homogeneous	two faster redox steps replace a slow reaction between two negatively charged ions; the iron ion is regenerated

For the iron-ion cycle:

S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺ 2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂

Adding the steps cancels the iron ions and gives the overall equation. That cancellation is the clearest check that the catalyst is regenerated.

Enzyme Activity vs pH (Typical Shape). Enzyme activity plotted as Reaction rate against pH. — Typical enzyme trend: highest activity near an optimum pH, with lower activity at more acidic or more alkaline conditions (shape and optimum vary by enzyme).

Values for Enzyme Activity vs pH (Typical Shape)
pH (unitless)	Enzyme activity
2	0
4	0.2
6	0.8
7	1
8	0.8
10	0.2
12	0

Values for Enzyme Activity vs pH (Typical Shape)
pH (unitless)	Enzyme activity
2	0
4	0.2
6	0.8
7	1
8	0.8
10	0.2
12	0

Worked Examples

Problem

Explain why increasing the surface area of a solid catalyst increases reaction rate.

Study the worked solution

Increase site availability
Method
State that a larger exposed surface provides more active sites.
Reason
Only accessible catalyst surface can adsorb reactant particles.
Working
Surface area ↑ ⇒ available active sites ↑.
Increase adsorption and reaction
Method
State that more reactant particles can adsorb simultaneously.
Reason
Adsorption brings particles onto the lower-energy surface pathway where bonds can weaken and rearrange.
Working
More adsorbed reactants can undergo surface reaction per unit time.
Conclude
Method
Link the larger number of surface events to rate.
Reason
Products desorb and free sites for further cycles.
Working
Successful catalytic events per unit time increase, so rate increases.

Problem

Explain why adding a catalyst to a reversible reaction at fixed temperature does not change its equilibrium position.

Try this before viewing the solution

Forward/reverse barriers, rates and equilibrium conclusion

Hints

Hint 1: treat both directions

The alternative pathway lowers activation energy for forward and reverse reactions.

Hint 2: separate speed from position

Both directions reach their equal-rate equilibrium sooner, but reaction energetics and temperature are unchanged.

View solution step by step

Describe the pathway
Method
State that the catalyst lowers activation energy for both directions.
Reason
The same alternative pathway connects reactants and products in either direction.
Working
Forward rate ↑ and reverse rate ↑.
Preserve equilibrium
Method
State that equilibrium is reached faster but at the same composition.
Reason
The catalyst changes neither temperature nor the relative energies determining K.
Working
Equilibrium position and K remain unchanged.

Learner claim

A learner says, “Any particle that hits an active site must react, because the catalyst removes the activation-energy barrier.” Identify both errors.

Classify the barrier

A catalyst usually

Lowers the activation barrierRemoves the barrier completely

View solution step by step

Correct the energy claim
Method
State that the alternative pathway has lower, not zero, activation energy.
Reason
Adsorbed particles still need sufficient energy for the surface reaction steps.
Working
A lower barrier increases successful fraction; it does not guarantee success.
Correct the collision claim
Method
State that adsorption geometry, energy and site availability still matter.
Reason
Some encounters fail to form the required bonds or occur at blocked/unsuitable sites.
Working
More active sites raise the event frequency, but not every collision reacts.

Problem

For steps

H₂O₂ + I⁻ → H₂O + IO⁻ and

H₂O₂ + IO⁻ → H₂O + O₂ + I⁻, identify the catalyst and intermediate and justify each. [4 marks]

Try this before viewing the solution

Species identities and step evidence

View solution step by step

Track iodide
2 marks
Method
Identify I⁻ as the catalyst.
Reason
It is consumed in the first step and regenerated in the second, so it cancels from the overall equation.
Working
Catalyst: I⁻.
Track hypoiodite
2 marks
Method
Identify IO⁻ as the intermediate.
Reason
It is formed in the first step and consumed in the second, so it also cancels but was not present initially.
Working
Intermediate: IO⁻.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Credit each identity and its consumed/formed/regenerated evidence.

Mark points

1 mark: Identifies iodide as the catalyst.
1 mark: Explains that iodide is consumed then regenerated.
1 mark: Identifies hypoiodite as the intermediate.
1 mark: Explains that hypoiodite is formed then consumed.

Problem

For each process below, name a suitable catalyst or catalytic material, classify the catalysis as homogeneous or heterogeneous, and give the key mechanistic idea: (i) the Haber process; (ii) a catalytic converter; (iii) oxidation of atmospheric sulfur dioxide; (iv) reaction of

S₂O₈²⁻ with

I⁻ using the

Fe²⁺/Fe³⁺ cycle. [6 marks]

Try this before viewing the solution

Catalyst, type and mechanistic reason for each process

View solution step by step

Haber process
1 mark
Method
Name iron and classify it as a heterogeneous catalyst.
Reason
Nitrogen and hydrogen adsorb on the solid surface, bonds weaken and ammonia eventually desorbs.
Working
N₂ and H₂ are gases; Fe is a solid surface.
Catalytic converter
1 mark
Method
Name platinum, palladium or rhodium surfaces and classify the catalysis as heterogeneous.
Reason
Exhaust gases adsorb and react on the metal surface to form less harmful products such as CO₂ and N₂.
Working
Gas reactants + solid catalyst = heterogeneous catalysis.
Atmospheric sulfur dioxide
1 mark
Method
Identify the homogeneous NO/NO₂ cycle.
Reason
NO is oxidised to NO₂; NO₂ then oxidises SO₂ to SO₃ and regenerates NO.
Working
2NO + O₂ → 2NO₂
NO₂ + SO₂ → NO + SO₃ (combine the steps in the correct ratio to cancel the nitrogen oxides).
Iron-ion cycle
3 marks
Method
Classify the cycle as homogeneous and show that the iron ions are regenerated.
Reason
All reacting species are in the same aqueous phase, and two faster electron-transfer steps avoid the slow direct collision between two negatively charged ions.
Working
S₂O₈²⁻ + 2Fe²⁺ → 2SO₄²⁻ + 2Fe³⁺
2Fe³⁺ + 2I⁻ → 2Fe²⁺ + I₂
The iron species cancel when the steps are added.

Self-mark with the mark scheme

Compare your response with each mark point. Select a point only when your response contains that evidence.

Credit accurate examples and phase classifications, plus regeneration and the two-step reason for the iron-ion cycle.

Mark points

1 mark: Names iron, heterogeneous catalysis and the surface route.
1 mark: Names Pt, Pd or Rh and the gas-on-solid surface route.
1 mark: Identifies the homogeneous NO/NO2 cycle and regeneration in sulfur dioxide oxidation.
1 mark: Classifies the iron-ion cycle as homogeneous.
1 mark: Shows Fe2+/Fe3+ conversion and regeneration.
1 mark: Explains the lower-energy two-step route or avoidance of the slow anion–anion reaction.

Next step

Cover the catalyst names, then reconstruct the reason for each one from phases, adsorption or regeneration rather than memorising four disconnected facts.

Problem

Explain why an enzyme rate–temperature graph rises to an optimum and then falls sharply.

Try this before viewing the solution

Low-temperature rise, optimum and high-temperature fall

Hints

Hint 1: before the optimum

Higher kinetic energy increases collision frequency and productive enzyme–substrate encounters.

Hint 2: after the optimum

Excessive heat disrupts the bonding that maintains the active-site shape.

View solution step by step

Explain the initial rise
Method
Link increasing temperature to faster-moving enzyme and substrate particles.
Reason
Collision frequency and productive complex formation increase up to the optimum.
Working
Temperature ↑ ⇒ rate ↑ initially.
Explain the sharp fall
Method
State that excessive temperature denatures the enzyme and alters its active site.
Reason
Fewer substrate molecules bind in a productive geometry.
Working
Above optimum: productive complexes ↓ ⇒ rate falls sharply.

Next step

Try the catalyst-poisoning Mind Stretcher independently.

Common Mistakes

Saying catalysts change Δ H or equilibrium position (they don’t).
Confusing denaturation with ‘slowing down’; denaturation changes enzyme shape/active site.
Forgetting to name what the catalyst does (alternative pathway, lower Eₐ).
For heterogeneous catalysts, describing “more collisions” without mentioning adsorption/active sites.

Use the Reaction Kinetics topic check to practise and check your understanding.

Exam Tips

Catalysts provide an alternative pathway with lower Eₐ; they are not consumed.
For enzymes, mention optimum temperature/pH and denaturation at extremes.
If asked for a graph, label axes and describe the trend (rise to optimum then fall).

Mind Stretchers

Mind stretcher 1Extension

In heterogeneous catalysis, explain why catalyst “poisoning” (a substance binding strongly to the surface) decreases reaction rate.

Show Hint

A catalyst changes the pathway and is regenerated; surface availability can therefore control its effectiveness.

Show Answer

Mark scheme:

Poison molecules occupy active sites on the catalyst surface.
Fewer reactant particles can adsorb and react.
Therefore the number of successful surface reactions per unit time decreases, lowering the rate.

Mind stretcher 2: Diagnosing catalyst poisoningExtension

Question. A solid catalyst works initially but slows sharply when a trace impurity is added, although temperature and reactant concentrations are unchanged. Give a particle-level explanation.

Show Hint

Consider adsorption sites rather than activation energy of the original uncatalysed route.

Show Answer

The impurity can adsorb strongly on active sites and block reactant adsorption. Fewer surface encounters follow the catalytic pathway, so the observed rate falls even though the remaining sites still provide the same lower-energy route.

Syllabus and review details

GCE A-Level H2 Chemistry 9476-2027 · 9476-2027
9476 (2027), complete syllabus

Last reviewed: 2026-08-29