Alternative Reaction Pathways and Activation Energy
Catalysts increase the rate of reaction by providing an alternative reaction pathway with a lower activation energy (\(E_{\text{a}}\)). They remain chemically unchanged at the conclusion of the reaction and do not alter the enthalpy change (\(\Delta H\)) or equilibrium constant (\(K\)).
How Catalysts Work
A catalyst speeds up a chemical reaction without being permanently changed. It does this by offering a completely different reaction mechanism that has a lower activation energy barrier. Because more particles now have sufficient energy to react, the rate increases.
Energy Profile: With and Without Catalyst
Key Points About Catalysts
| Feature | Effect |
|---|---|
| Activation energy | Lowered (provides alternative pathway) |
| Enthalpy change (ΔH) | Unchanged (same reactants and products) |
| Rate of reaction | Increased (both forward AND reverse) |
| Position of equilibrium | Unchanged (equilibrium is reached faster, but K stays the same) |
| Catalyst consumed? | No (regenerated at the end of the mechanism) |
Maxwell-Boltzmann Distribution
A catalyst does not change the shape of the Maxwell-Boltzmann curve or the average kinetic energy of particles. Instead, it shifts the activation energy threshold to the left, meaning a greater proportion of particles now have sufficient energy to react.
Maxwell-Boltzmann: Effect of a Catalyst
Biological Catalysts: Enzymes
Biological catalysts are known as enzymes. Note that the IB Chemistry syllabus does not require memorisation of complex biological enzyme mechanisms; understanding the general catalytic principle (lowering \(E_{\text{a}}\) via active sites) is sufficient.
Catalysts Do Not Lower the Existing Energy Barrier
Examiner Trap: Never state that a catalyst "lowers the activation energy of the existing pathway". A catalyst provides a completely new, alternative pathway (often involving intermediates) with its own intrinsically lower activation energy barrier.
Why Catalysts Do Not Shift Equilibrium Position
A catalyst speeds up the forward and reverse reaction rates by the exact same factor (lowering \(E_{\text{a}(\text{forward})}\) and \(E_{\text{a}(\text{reverse})}\) by the same amount). Therefore, equilibrium is reached much faster, but the equilibrium position and the equilibrium constant \(K\) remain completely unchanged.
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