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How do catalysts speed up reactions without being used up?

Explain how a catalyst increases rate by providing an alternative pathway with lower activation energy.

How catalysts lower activation energy, the difference between homogeneous and heterogeneous catalysts, enzymes, and the effect on the energy profile and Maxwell-Boltzmann distribution, with worked TASC-style examples.

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What this dot point is asking

TASC expects you to explain the mechanism by which a catalyst works, describe its effect on the energy profile and the Maxwell-Boltzmann distribution, and distinguish catalyst types.

What a catalyst does

A catalyst increases the rate of a reaction without being permanently used up. It takes part, often forming an intermediate, but is regenerated by the end so the same catalyst molecule acts again and again. Because it is not consumed, only a small amount is needed.

The lower activation energy pathway

The key idea is that a catalyst lowers the activation energy. On an energy profile diagram the catalysed pathway has a lower peak (energy barrier) than the uncatalysed pathway, while the energies of the reactants and products are unchanged. Because the barrier is lower, a greater proportion of colliding particles have enough energy to react.

On a Maxwell-Boltzmann distribution, lowering EaE_a moves the EaE_a line to the left, so the area to its right (the fraction of particles that can react) is much larger. The temperature is unchanged, so the curve itself does not move.

Types of catalyst

  • Homogeneous catalysts are in the same physical state as the reactants, typically all in solution, and usually work by forming an intermediate species.
  • Heterogeneous catalysts are in a different state, commonly a solid with gaseous or liquid reactants. They adsorb reactant molecules onto active sites on their surface, weakening bonds and holding molecules in a favourable orientation; examples include iron in the Haber process and platinum or palladium in catalytic converters.
  • Enzymes are biological catalysts (large protein molecules) that catalyse reactions in living things with great specificity, each having an active site shaped to fit a particular substrate.

Catalysts in industry

Catalysts are central to industrial chemistry because they let reactions run at lower temperatures and pressures, saving energy and cost. Iron catalyses the Haber process for ammonia, vanadium(V) oxide catalyses the Contact process for sulfuric acid, and platinum, palladium and rhodium in a vehicle's catalytic converter speed the conversion of toxic exhaust gases (carbon monoxide and nitrogen oxides) into less harmful carbon dioxide and nitrogen. Because a catalyst lets a process reach an acceptable rate at a lower temperature, it can also improve the equilibrium yield indirectly: for an exothermic reaction, a lower temperature gives a higher equilibrium yield, and the catalyst restores the rate that the lower temperature would otherwise cost.

Catalyst poisoning

A catalyst can be deactivated by a poison, a substance that binds strongly to the active sites and blocks reactant molecules. Lead poisons the platinum in catalytic converters, which is why leaded petrol had to be phased out. Poisoning explains why catalysts, though not consumed by the reaction itself, may still need periodic replacement in real plants.

In the exam, always state that the catalyst provides an alternative pathway with lower activation energy, and stress that it is regenerated and does not alter the enthalpy change or the equilibrium position.

Exam-style practice questions

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

TCE 20233 marksThe reaction 4NH3(g)+5O2(g)4NO(g)+6H2O(g)4\text{NH}_3(g) + 5\text{O}_2(g) \rightarrow 4\text{NO}(g) + 6\text{H}_2\text{O}(g) usually takes place in the presence of a platinum-rhodium catalyst. Using a molecular-energy distribution diagram and collision theory, explain how a catalyst speeds up a reaction.
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A catalyst provides an alternative reaction pathway with a lower activation energy EaE_a. On the Maxwell-Boltzmann distribution this is shown by drawing the catalysed EaE_a line further to the left of the uncatalysed EaE_a. (1 mark)

Because the catalysed EaE_a is lower, a larger fraction of the molecules (a greater area under the curve to the right of EaE_a) now have enough energy to react on collision. (1 mark)

So at the same temperature a greater proportion of collisions are successful per unit time, which increases the reaction rate. The catalyst itself is not consumed. (1 mark)

TCE 20222 marksIndustrially, a vanadium(V) oxide catalyst is used to increase the rate of the conversion 2SO2(g)+O2(g)2SO3(g)2\text{SO}_2(g) + \text{O}_2(g) \rightleftharpoons 2\text{SO}_3(g). Outline how a catalyst changes the rate of a chemical reaction and state its effect on the position of equilibrium.
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A catalyst speeds up a reaction by providing an alternative pathway (mechanism) with a lower activation energy than the uncatalysed reaction. Because EaE_a is lower, a greater proportion of colliding molecules have enough energy to react, so the number of successful collisions per unit time increases and the rate rises. (1 mark)

The catalyst lowers EaE_a for the forward and reverse reactions equally, so it reaches equilibrium faster but does not change the position of equilibrium or ΔH\Delta H; it is regenerated and not used up overall. (1 mark)

TCE 20212 marksA statement found online claims that UV light acts as a catalyst to speed up the decomposition of hydrogen peroxide into water and oxygen. Justify that this statement is incorrect, referring to what is meant by a catalyst.
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A catalyst is a substance that increases the rate of a reaction by providing an alternative pathway with a lower activation energy, and is regenerated (not consumed) overall. (1 mark)

UV light is a form of energy, not a substance, so it cannot be a catalyst. UV speeds the decomposition by supplying energy to the reactant molecules (photochemical activation), helping them reach the activation energy, rather than by lowering EaE_a. It therefore does not fit the definition, so the statement is incorrect. (1 mark)

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