IB Chemistry SLTopic 5 — The Rate of ReactionPaper 1 & 2Core idea~12 min read
Energy Profiles With and Without Catalysts
A catalyst does not push the reaction harder. It finds a different way round — a route over a lower pass — and because the barrier is lower, far more of the collisions already happening are now enough.
📚 What you need to know
A catalyst increases the rate by providing an alternative pathway with a lower activation energy.
It is chemically unchanged at the end of the reaction and is not used up.
It does not change ΔH, the energies of the reactants or products, or the amount of product obtained.
It lowers Ea for the forward and reverse reactions by the same amount.
Homogeneous catalysts are in the same phase as the reactants; heterogeneous catalysts are in a different phase.
Industrially, catalysts allow lower temperatures and pressures and improve selectivity.
Two routes, one destination
Same starting level, same finishing level. Only the height of the pass between them differs.
Read what has and has not moved. The reactants start at the same energy. The products finish at the same energy. Therefore ΔH is identical for both routes — a catalyst cannot make a reaction more exothermic, and it cannot change how much product you eventually get.
What has changed is the barrier. With a lower Ea, a greater proportion of collisions carry enough energy to react, so more successful collisions happen per second and the rate rises.
Notice the reverse arrow too. Because both routes end at the same peak, the catalyst lowers Ea for the backward reaction by exactly the same amount. A reversible reaction therefore reaches equilibrium sooner, but at exactly the same position — a favourite exam point.
WORKED EXAMPLE
Using the profile above, state Ea for the uncatalysed and catalysed routes and the value of ΔH. Then calculate Ea for the reverse reaction by each route.
Step 1 — read the levelsReactants 100, uncatalysed peak 200, catalysed peak 155, products 45 kJ mol⁻¹.Step 2 — forward activation energiesuncatalysed: 200 − 100 = 100catalysed: 155 − 100 = 55Step 3 — enthalpy change45 − 100 = −55ΔH = −55 kJ mol⁻¹ for BOTH routesStep 4 — reverse activation energiesuncatalysed: 200 − 45 = 155catalysed: 155 − 45 = 110Both reverse values are 45 lower, exactly as both forward values were. The catalyst dropped the peak by 45 and that is all it did.
Homogeneous and heterogeneous
Homogeneous
Heterogeneous
Phase
Same phase as the reactants
Different phase from the reactants
Typical example
An acid catalysing a reaction between two solutions
Solid iron in the Haber process, with gaseous reactants
How it works
Forms an intermediate that then breaks down, releasing the catalyst
Reactants adsorb onto the surface, react there, then desorb
Practical note
Mixes intimately, but must be separated from the products afterwards
Easy to separate and reuse; performance depends on surface area
The surface does two jobs: it holds the molecules together in the right orientation, and it weakens their bonds so less energy is needed to break them.
Because a heterogeneous catalyst only works at its surface, it is usually made into a fine mesh, a gauze or a coating on a porous support — maximising surface area for the smallest quantity of what is often an expensive metal.
Where catalysts come from
Enzymes are biological catalysts, extraordinarily specific because the substrate must fit an active site. They let living cells and industrial processes run reactions at low temperature and near-neutral pH that would otherwise need harsh conditions.
Transition metals are widely used because their variable oxidation states let them accept and release electrons, providing alternative routes for redox reactions. Iron, nickel, platinum and vanadium(V) oxide are the standard industrial examples.
The environmental case for catalysts is worth being able to state. They allow lower temperatures and pressures, so less energy is used and less CO2 is released generating it. They improve selectivity, suppressing side reactions, which means fewer by-products and a higher atom economy. And being unchanged, small amounts can be recovered and reused indefinitely.
WORKED EXAMPLE
A reaction has Ea = 75 kJ mol–1 and ΔH = –40 kJ mol–1. A catalyst reduces Ea to 45 kJ mol–1. Calculate the reverse activation energy with and without the catalyst, and state the effect on the yield.
Step 1 — uncatalysed reverseEₐ(rev) = 75 − (−40) = 115 kJ mol⁻¹Step 2 — catalysed reverseEₐ(rev) = 45 − (−40) = 85 kJ mol⁻¹Step 3 — compareboth lowered by 30 kJ mol⁻¹Step 4 — effect on yieldNone. ΔH is unchanged, both directions are sped up equally, so the equilibrium position and the final amount of product are exactly the same — reached sooner.
💡 Exam tip
The mark-scheme phrase is “provides an alternative pathway of lower activation energy”. Use it.
When sketching, draw the catalysed curve with the same reactant and product levels and only a lower peak.
State explicitly that ΔH is unchanged — questions frequently award a mark for that alone.
Say chemically unchanged rather than “not involved”. A catalyst does take part; it is simply returned.
For evaluation marks, link catalysts to lower energy use, better selectivity and higher atom economy.
⚠️ Common mix-up
Drawing the catalysed curve with lower products, which wrongly changes ΔH.
Saying a catalyst lowers the activation energy of the reaction. It offers a route with a lower one.
Claiming a catalyst increases the yield. It changes the time taken, not the amount.
Thinking only the forward reaction is catalysed. Both directions are, equally.
Mixing up homogeneous and heterogeneous. Same phase, different phase — check the state symbols.
Up next: Maxwell–Boltzmann Distributions — the graph that finally shows you the “greater proportion of particles” that every explanation in this sub-topic has been referring to.
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