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SAChemistrySyllabus dot point

How does a catalyst speed up a reaction without being consumed?

Explain how catalysts increase reaction rate by providing an alternative pathway with a lower activation energy, and represent this on an energy profile.

How catalysts lower activation energy by offering an alternative pathway, the Maxwell-Boltzmann explanation, homogeneous versus heterogeneous catalysis, energy profiles, and worked SACE-style rate and energy calculations.

Generated by Claude Opus 4.88 min answerUpdated 2026-06-02

Reviewed by: AI editorial process; not yet individually human-reviewed

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What this dot point is asking
Lead worked calculation
How a catalyst works
The energy profile
Homogeneous and heterogeneous catalysts
What a catalyst does not change
Why it matters for managing processes

What this dot point is asking

SACE expects you to explain catalysis with collision theory, draw and label energy profiles showing the lowered $E_a$ , distinguish homogeneous and heterogeneous catalysts, and state what a catalyst does and does not change.

Lead worked calculation

Comparing successful-collision fractions

The uncatalysed decomposition of a compound has $E_a = 80\ \text{kJ mol}^{-1}$ ; a catalyst lowers it to $50\ \text{kJ mol}^{-1}$ . Explain quantitatively, in words, why this dramatically increases the rate.

Identify what controls rate

Rate is proportional to the fraction of molecules with energy at least $E_a$ . On the Maxwell-Boltzmann curve this is the area to the right of the $E_a$ line.

Compare the two thresholds

At $E_a = 80\ \text{kJ mol}^{-1}$ , only a small tail of molecules has enough energy. Moving the threshold left to $E_a = 50\ \text{kJ mol}^{-1}$ captures a much larger area under the curve.

Conclude

Because the number of molecules able to react grows steeply as the threshold falls, even a $30\ \text{kJ mol}^{-1}$ reduction can increase the successful-collision fraction many times over, so the rate increases sharply, while the products and $\Delta H$ stay the same.

How a catalyst works

The catalyst does not give the particles more energy; it lowers the energy barrier they must clear. With a lower $E_a$ , a greater proportion of the existing collisions already have enough energy to succeed, so the rate of successful collisions, and thus the reaction rate, increases.

The energy profile

An energy profile (reaction-coordinate diagram) plots energy against the progress of the reaction.

Homogeneous and heterogeneous catalysts

Homogeneous catalysts are in the same phase as the reactants (e.g. aqueous ions catalysing a solution reaction). They react to form an intermediate, then are regenerated.
Heterogeneous catalysts are in a different phase, usually a solid catalysing gaseous or liquid reactants (e.g. iron in the Haber process, platinum in catalytic converters). Reactants adsorb onto the catalyst surface, which weakens their bonds and holds them in favourable orientations; the products then desorb, freeing the surface.

What a catalyst does not change

A catalyst lowers $E_a$ for both the forward and reverse reactions equally, so it speeds the approach to equilibrium but does not shift the equilibrium position or change $K_c$ . It also does not change the enthalpy change $\Delta H$ , the amount of product at equilibrium, or the identity of the products. It only changes how fast equilibrium is reached.

Why it matters for managing processes

Catalysts let industry run reactions faster at lower temperatures and pressures, saving energy and cost, which is why they are central to green and economic process design. Roughly $90\%$ of industrial chemical processes use a catalyst, from the iron in ammonia synthesis to the metals in vehicle catalytic converters that destroy pollutants.

Exam-style practice questions

Practice questions written in the style of SACE Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

SACE 20224 marksAn uncatalysed reaction has an activation energy of

75\ \text{kJ mol}^{-1}

and a catalysed pathway has an activation energy of

45\ \text{kJ mol}^{-1}

. (a) Sketch and label an energy profile showing both pathways for this exothermic reaction. (b) Using the Maxwell-Boltzmann distribution, explain why the lower activation energy increases the rate. (c) State the effect of the catalyst on the enthalpy change of the reaction.

Show worked answer →

(a) The profile shows reactants higher than products (exothermic, $\Delta H$ negative). Two humps are drawn: the uncatalysed peak at $+75\ \text{kJ mol}^{-1}$ above reactants and a lower catalysed peak at $+45\ \text{kJ mol}^{-1}$ . Both start and end at the same reactant and product levels. (2 marks)

(b) On the Maxwell-Boltzmann distribution, lowering $E_a$ moves the $E_a$ line to the left, so a larger fraction of particles (a greater area under the curve) now has energy at least $E_a$ . More collisions are successful per second, so the rate increases. (1 mark)

(c) The catalyst has no effect on $\Delta H$ . It changes the pathway and the activation energy but not the energies of the reactants or products, so the enthalpy change is unchanged. (1 mark)

SACE 20203 marksHydrogen peroxide decomposes slowly:

2\text{H}_2\text{O}_2 \rightarrow 2\text{H}_2\text{O} + \text{O}_2

. Adding manganese(IV) oxide greatly speeds it up, and the

\text{MnO}_2

can be recovered unchanged afterwards. (a) Calculate the mass of oxygen produced from

3.40\ \text{g}

\text{H}_2\text{O}_2

. (b) Explain why the

\text{MnO}_2

is described as a catalyst. (

M(\text{H}_2\text{O}_2) = 34.0

M(\text{O}_2) = 32.0\ \text{g mol}^{-1}

Show worked answer →

(a) $n(\text{H}_2\text{O}_2) = \dfrac{3.40}{34.0} = 0.100\ \text{mol}$ . From the $2:1$ ratio, $n(\text{O}_2) = 0.0500\ \text{mol}$ , so $m(\text{O}_2) = 0.0500 \times 32.0 = 1.60\ \text{g}$ . (2 marks)

(b) $\text{MnO}_2$ increases the rate by providing an alternative pathway with a lower activation energy, yet is chemically unchanged and recoverable at the end, so it is a catalyst. (1 mark)