The big idea: A reaction only happens when particles collide — and only a successful (effective) collision causes a reaction. An effective collision needs two things at once:
- enough energy — at least the activation energy, Ea, and - the correct orientation (the right parts of the particles must line up).
So to speed a reaction up you either make collisions happen more often, or make a larger fraction of them successful. Five factors do this.
Two ways to raise the rate: Every rate factor works through one or both of these:
- Collision frequency — how often particles meet (concentration, pressure, surface area, temperature). - Energy of the collisions — what fraction clear the Ea barrier (temperature, catalyst).
Temperature is special: it does both. Concentration/pressure/surface area change only the frequency; a catalyst changes only the fraction that succeed (by lowering Ea).
Define your terms: - Activation energy (E_{a}) — the minimum energy a colliding pair must have for a reaction to occur. - Effective (successful) collision — one with energy ≥ Ea and the correct orientation. - Catalyst — a substance that speeds up a reaction by providing an alternative pathway of lower E_{a}, and is not used up itself.
These three factors all work the same way: they pack more reactant particles into the space where the reaction happens, so the particles collide more often. More frequent collisions means more successful collisions each second — a faster rate.
Lower rate
- dilute solution / low gas pressure
- a solid in large lumps (small surface area)
- particles are spread out → collisions are infrequent
Higher rate
- concentrated solution / high gas pressure
- a solid as a fine powder (large surface area)
- particles are crowded together → collisions are frequent
Same idea, three settings: - Concentration: doubling the concentration of an acid roughly doubles the collision frequency with a metal. - Pressure: raising the pressure of a gas mixture pushes the molecules closer, exactly like concentrating a solution. - Surface area: powdered calcium carbonate fizzes far faster than a marble chip because much more of it is exposed to the acid.
In every case the average energy of the particles is unchanged — only how often they meet.
Watch the wording: These three do not make collisions more energetic and they do not change the activation energy. For the explain mark you must say collision frequency increases — not 'the particles have more energy'.
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Not all particles in a sample move at the same speed. The Maxwell-Boltzmann distribution shows how the kinetic energies are spread out: a few particles are slow, a few are very fast, and most are somewhere in the middle. Only the particles to the right of the Ea line have enough energy to react.
Raising the temperature shifts the curve to the right and lower (the area under it is unchanged). Eₐ is fixed, so a much larger fraction of molecules now have energy ≥ Eₐ — that shaded fraction can react.
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Why temperature has such a big effect: Raising the temperature does two things:
- the particles move faster, so they collide more often (more frequent collisions), and - the whole distribution shifts to the right, so a much larger fraction of particles now have energy ≥ Ea.
The second effect is the main reason: even a small temperature rise moves many more particles past Ea, so the proportion of successful collisions rises sharply. That is why rates roughly double for a ~10 °C rise.
A catalyst works completely differently. It does not heat the particles or change the distribution. Instead it provides an alternative pathway with a lower activation energy.
A catalyst lowers the activation energy (Eₐ → Eₐ(cat.)). At the SAME temperature, a larger fraction of molecules now have enough energy to react.
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A catalyst provides an alternative route with a lower activation energy (dashed). The reactant and product levels — and therefore ΔH — are unchanged.
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Catalyst: lower the bar, not the runners: The distribution curve stays exactly the same — but the Ea line moves left. So at the same temperature, a larger fraction of molecules already have energy ≥ Ea(cat.). The catalyst is not consumed, and because the reactant and product energies are unchanged, ΔH is unchanged too.
Worked example — explaining a temperature rise
Explain, using the Maxwell-Boltzmann distribution, why raising the temperature speeds up a reaction.
Solution
- At a higher temperature the particles have a greater average kinetic energy and move faster, so they collide more frequently.
- More importantly, the distribution shifts to the right, so a greater proportion of particles have energy ≥ E_{a}.
- Therefore a larger fraction of collisions are successful (energy ≥ Ea and correct orientation), so the rate increases.
Final answer
Faster + more frequent collisions, and a larger fraction with energy ≥ Ea, so more successful collisions per second.
How this is tested: R2.2 rate factors are a banker on every paper, in two flavours.
- Paper 1A (MCQ): which changes increase the rate?, or identify the lower-temperature curve / the catalysed E_{a} on a distribution or profile. - Paper 2: a 2-3 mark explain — usually why does raising temperature increase the rate? or how does a catalyst increase the rate?, sometimes asking you to annotate a Maxwell-Boltzmann curve.
The distinction the markers reward: temperature changes both frequency and the fraction with energy ≥ Ea; a catalyst lowers E_{a} without changing the distribution.
Score the explain marks: (1) Always name collision frequency and (for temperature/catalyst) the fraction with energy ≥ E_{a}. (2) For a catalyst say 'alternative pathway of lower activation energy' — and that ΔH is unchanged. (3) If asked to sketch a hotter curve, draw it lower and shifted right, with the same area underneath.
IB-style question — temperature and the distribution (a)
Hydrogen peroxide decomposes slowly at room temperature. (a) Explain, with reference to the Maxwell-Boltzmann distribution, why warming the solution increases the rate of decomposition. [3]
How to score the marks
- Mark 1 — frequency. At the higher temperature the particles have greater average kinetic energy and move faster, so they collide more frequently.
- Mark 2 — the distribution. The Maxwell-Boltzmann curve shifts to the right (it flattens and broadens), so a greater proportion of particles have energy ≥ E_{a}.
- Mark 3 — successful collisions. Therefore a larger fraction of collisions are successful (energy ≥ Ea), and this — more than the frequency change — increases the rate.
Final answer
More frequent collisions; the distribution shifts right so a greater fraction has energy ≥ Eₐ; hence more successful collisions per second.
IB-style question — adding a catalyst (b)
A small amount of solid manganese(IV) oxide is added to the hydrogen peroxide. (b) (i) Explain how the catalyst increases the rate of reaction. (ii) State two observations that would confirm the manganese(IV) oxide is acting as a catalyst. [4]
How to score the marks
- Mark 1 (i) — lower E_{a}. The catalyst provides an alternative reaction pathway with a lower activation energy.
- Mark 2 (i) — larger fraction reacts. So at the same temperature a greater proportion of molecules have energy ≥ Ea(cat.), giving more successful collisions and a faster rate.
- Mark 3 (ii) — observation 1. The reaction speeds up (faster fizzing / more gas per second / the reaction finishes sooner).
- Mark 4 (ii) — observation 2. The mass (and chemical nature) of the manganese(IV) oxide is unchanged at the end — it is recovered, showing it was not used up.
Final answer
(i) alternative pathway of lower Eₐ → larger fraction has energy ≥ Eₐ → faster. (ii) faster gas evolution AND the solid is recovered unchanged at the end.