From C6 Rate & Equilibrium you already know that particles must collide to react, and that temperature, concentration, surface area and catalysts speed reactions up. A-Level asks why — and answers with one idea: a collision only counts if the particles meet with at least the activation energy and in the right orientation. That single bar — cleared or not — explains every factor on this page, and the Maxwell–Boltzmann distribution is the tool that lets you see what fraction of collisions clear it.
Collision theory & activation energy
Chemical reactions happen when particles collide — but a collision on its own is not enough. Most collisions bounce apart unchanged. For a collision to lead to a reaction, two things must be true at once:
- the particles must collide with at least the activation energy — enough energy to start breaking bonds;
- they must collide in the correct orientation — the right parts of each particle facing each other.
Fail either test and the particles simply rebound. At ordinary temperatures the energy test is the one most collisions fail — which is why reactions take the time they do, and why every lever later on this page works by changing how often particles collide, or what fraction of those collisions succeed.
Activation energy (Ea) — the minimum energy that colliding particles must have for a reaction to occur.
That minimum has a picture: it is the height of the barrier on the reaction profile — the same hump you drew in Energetics, named there and parked for here. Particles sitting at the reactant level have to be pushed up and over it before they can roll down to products, and the activation energy is how big that push must be.
How reactive two chemicals are depends partly on themselves — on how high their activation-energy barrier is. But the factors on this page change the rate without touching the chemicals: they work on the collisions, or (for a catalyst) on the barrier. Keep that split in mind — it is the whole topic in one line.
Interactive — the collision sandbox: watch how few collisions succeed
Push the energy up and more collisions clear Ea — but even then, most collisions still fail. Switch orientation on and the odds fall again.
- “They collide” is never a complete reason. Name both requirements — enough energy (≥ Ea) and correct orientation. A collision that has one but not the other does not react.
- Activation energy is a minimum. Particles need energy equal to or greater than Ea — write “≥”, not just “greater than”.
- The factors change collisions, not the barrier. Temperature, concentration and pressure leave Ea exactly where it is — only a catalyst lowers it (and it does so by offering a different route, not by shrinking the original barrier).
GCSE groundwork: C6 — collision theory; the barrier itself sits on the profile in Energetics.
🧪 Exam-style questions
A mixture of 2 dm3 of hydrogen and 1 dm3 of oxygen is at room temperature. Which statement is correct? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Rate of reaction, and how you measure it
Before you can change a rate, you need to know what “rate” means and how to put a number on it.
Rate of reaction — the change in concentration of a reactant or product per unit time.
A rate is always “an amount of something, per unit of time”, so its units follow whatever you choose to measure: mol dm−3 s−1 if you track concentration, cm3 s−1 for a gas volume, g s−1 (or g min−1) for mass. Whatever you measure, the value is meaningless without its unit — hold on to that, it is where marks go.
Two experimental approaches cover almost every rate experiment you will meet:
- Continuous monitoring — follow one quantity all the way through: gas collected in a syringe, or mass lost on a balance as a gas escapes (a cotton-wool plug lets CO2 out but keeps spray in). You get a smooth curve: steepest at the start, flattening as the reactants run low.
- Clock reactions — time how long to reach one fixed, visible change (a cross disappearing behind cloudiness; a sudden colour). One run gives one average rate, and the shorter the time, the faster the reaction: rate ∝ 1/time.
Interactive — drag the line until it sits as the tangent at 2 minutes
A gas is collected in a syringe. To find the rate at exactly 2 minutes, draw a tangent to the curve at that point and read its gradient from two points that sit on the tangent line.
Step 1 — pick two clear points on the tangent line:
(60 s, 44 cm3) and (180 s, 80 cm3)
Step 2 — gradient = change in volume ÷ change in time:
rate = (80 − 44) ÷ (180 − 60) = 36 ÷ 120
Step 3 — state the value with its unit (this is the marked step):
rate = 0.30 cm3 s−1
Read the tangent’s own points, not points on the curve — and never quote a bare “0.30”. A gradient without a unit scores nothing.
The other classic is the iodine clock: the mixture stays colourless, then turns blue-black the moment free iodine appears with the starch indicator. Naming the correct change — and which way round it goes — is exactly what examiner reports found students could not do.
Everything quantitative that grows out of these graphs — how the rate depends on each concentration, orders of reaction and the rate constant — is Year 2 work, waiting in rate equations.
- Every rate carries a unit. Examiner reports flag gradients quoted with no unit (or a malformed one) — write mol dm−3 s−1, cm3 s−1, g min−1, whatever fits what you measured.
- Instantaneous rate needs a tangent. Read the slope of a tangent to the curve, not a chord between two far-apart points.
- Describe the specific change. For a clock reaction, say what is actually seen — a cross hidden by cloudy sulfur, or colourless → blue-black — not a vague “colour change”.
- Read the stem for the measured variable. AQA often sets familiar rate chemistry in unfamiliar apparatus; check what is being measured before you calculate.
GCSE groundwork: C6 — calculating rates of reaction.
🧪 Exam-style questions
Hydrogen peroxide solution decomposes to form water and oxygen, 2H2O2(aq) → 2H2O(l) + O2(g), catalysed by manganese(IV) oxide. A student follows the reaction by collecting the oxygen in a gas syringe. Explain why the reaction is fastest at the start.
Show answer
At the start, the concentration of hydrogen peroxide is at its highest. 1 mark
So the reactant particles collide most frequently, giving the greatest frequency of successful collisions per unit time. 1 mark
As the H2O2 is used up its concentration falls, collisions become less frequent, and the rate drops — which is why the curve levels off.
The graph in Figure 1 shows how the concentration of hydrogen peroxide changes with time in the experiment from Q1. Tangents to the curve can be used to determine rates of reaction. Draw a tangent to the curve when the concentration of hydrogen peroxide solution is 0.05 mol dm−3. Use your tangent to calculate the gradient of the curve at this point.
Show answer
Tangent drawn with a ruler, touching the curve at 0.05 mol dm−3 (≈ 20 s) without crossing it. 1 mark
gradient = (0.014 − 0.075) ÷ (50 − 0) = −1.2 × 10−3 mol dm−3 s−1
Any value from −0.00120 to −0.00155 scores; the sign is ignored. 1 mark
The tangent must touch the curve at 0.05 mol dm−3 and not cross it — if white space shows between line and curve at the touch point, it crosses. On paper you draw the tangent yourself; read two well-separated points on the tangent, not on the curve.
A student investigated the reaction between magnesium and hydrochloric acid, Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g). The results are plotted on this graph. Which value is closest to the rate of reaction, in cm3 s−1, at 70 s? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
The Maxwell–Boltzmann distribution
The particles in a gas do not all move at the same speed, so they do not all have the same energy. At any instant a few are almost stationary, a few are moving very fast, and most are somewhere in between. The Maxwell–Boltzmann distribution is the shape of that spread — and it is the single most useful picture in kinetics, because it shows you the fraction of particles that can react.
Plot the number of molecules against energy and you get a lopsided hump. Its shape is not decorative — every feature is examined, and drawing it carelessly is a well-known way to lose marks.
- It starts at the origin (0, 0) — no molecule has zero energy.
- It rises to a peak at the most probable energy.
- The mean energy sits a little to the right of the peak (the curve is skewed).
- The tail approaches the x-axis but never touches it — there is no maximum energy.
- The total area under the curve = the total number of molecules.
Now overlay the activation energy as a vertical line. The particles that can react are those with E ≥ Ea — the area under the curve to the right of that line. At ordinary temperatures it is a thin sliver, which is precisely why most collisions fail (from collision theory) and why every lever on this page matters. Change the temperature or add a catalyst and you change the size of that sliver; that is the mechanism behind the two sections that follow.
Interactive — the Maxwell–Boltzmann explorer: heat it, catalyse it, watch the green area
The area under the curve never changes — the same molecules just spread to higher energies. Only the green slice beyond Ea decides the rate.
- Start at the origin; never touch the axis. A curve that begins above zero, or that meets the x-axis at high energy, loses the marks for those features every time.
- Label the axes the right way round. Energy on the x-axis, number of molecules on the y-axis — examiner reports flag reversed axes.
- The reactive population is an area, not a point. It is everything with E ≥ Ea — shade the region to the right of the line, do not just mark the line.
🧪 Exam-style questions
Which statement about the distribution curve of molecular energies in an ideal gas at a given temperature is correct? Tick (✓) one box.
Consider how the shape of the distribution-of-energies curve changes when the temperature of a gas is increased. Which is a correct statement about the gas molecules at a higher temperature? Tick (✓) one box.
A Maxwell–Boltzmann distribution is drawn for a gaseous reactant. What is represented by the total area under the curve? Tick (✓) one box.
The figure below shows the Maxwell–Boltzmann distribution of molecular energies in a sample of gas.
Label the y-axis on the figure above.
Show answer
Number of molecules (with a particular energy) 1 mark
“Amount”, “fraction” or “proportion” of molecules is allowed. “Particles” or “atoms” instead of molecules is ignored rather than penalised.
State why the curve starts at the origin.
Show answer
There are no molecules with no energy / all molecules have some energy. 1 mark
State what X indicates on the figure above.
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The most probable energy (the modal energy). 1 mark
Not the mean — the mean sits a little to the right of the peak.
Half of the gas molecules in the sample are removed. The remaining gas molecules are kept at the same temperature. Draw the new distribution of molecular energies for the remaining gas on the figure above.
Show answer
Peak at the same energy — in line with X (same temperature, so the shape’s position is unchanged). 1 mark
Area under the new curve is about half the original, and after leaving the origin the new curve never touches the old one. 1 mark
Source: AQA A-Level Chemistry past papers.
Temperature & Required Practical 3
Raising the temperature speeds almost every reaction up. There are two reasons, and you need to get the balance between them right:
- A minor effect: the particles move faster, so they collide slightly more often.
- The dominant effect: a greater proportion of particles now have energy ≥ Ea, so a greater proportion of collisions are successful.
Collision frequency barely changes with a modest temperature rise. What changes dramatically is the fraction of particles over the activation-energy bar — and that is what makes the reaction go faster. If you only ever remember one sentence from this page, make it that one.
That last point explains why a small temperature rise gives a large rate increase: the reactive particles live in the exponential tail, so nudging the whole distribution to the right pushes a disproportionately big extra slice past Ea. As a rough guide many reactions roughly double in rate for every 10 °C — useful intuition, not a law to quote.
“Explain why increasing the temperature increases the rate of reaction.” A full-mark answer reads:
Increasing the temperature increases the average kinetic energy of the particles. 1 mark A greater proportion of particles now have energy ≥ the activation energy. 1 mark So a greater proportion of collisions are successful, and the rate increases. 1 mark (Collision frequency also rises slightly — a real but minor effect that is usually not the marking point.)
The second sentence is the one students miss. “Particles move faster” or “they collide more often” on their own is not enough — the mark scheme wants the greater proportion with energy ≥ Ea.
The named method is the thiosulfate clock: to a fixed volume of sodium thiosulfate add a small measured volume of hydrochloric acid over a marked cross, and time how long the cloudiness (a fine precipitate of sulfur) takes to hide it. Repeat over about five temperatures, set with a water bath. Shorter time means faster reaction, so rate ∝ 1/time; plot 1/time against the mean temperature. Practical skills are examined across all three papers — see the required practicals page.
Almost every method mark here exists because the end-point is judged by eye, so everything except the temperature has to be pinned down:
- Change only the temperature. Fix every volume and concentration, and stand both solutions in the water bath to reach its temperature before you mix them, so the run starts at the temperature you record.
- Judge the end-point identically every time — same observer, same cross, and, easily missed, the same reaction vessel. A wider vessel spreads the mixture into a shallower layer, so you look down through less solution and the cross takes longer to vanish; change the vessel and the times no longer compare.
- Do not contaminate the concentrations. Rinse glassware with distilled water — never with the acid, whose residue would add to the next run and raise its concentration — and keep a separate measuring cylinder for each solution.
- Allow for temperature drift. The mixture’s own temperature shifts as it reacts, so read it at the start and end of each run and plot the mean, not the bath’s nominal value.
- Handle the sulfur dioxide. SO2 is toxic, so work in a well-ventilated room and tip each cloudy mixture straight into a sodium carbonate “stop bath” to neutralise the acid and the gas.
Interactive — build the mark-scheme answer
“Explain why increasing the temperature increases the rate of a reaction.” Select every statement that earns a mark — and nothing that doesn’t. Order doesn’t matter: AQA credits each point on its own.
your answer, assembled
- “Particles move faster” is not enough. Examiner reports flag this exact answer — you must add that a greater proportion of particles have energy ≥ Ea. That is the marking point.
- The successful fraction dominates, not collision frequency. Increased collisions is a real but minor effect — say so, and put the weight on the reactive fraction.
- The area is conserved. When temperature changes, the Maxwell–Boltzmann curve flattens and spreads — the total area (number of molecules) stays the same. It does not gain particles.
🧪 Exam-style questions
Increasing the temperature causes the rate of reaction to increase. Explain why a small increase in temperature causes a large increase in the rate of reaction.
Show answer
Many more particles now have energy equal to or greater than the activation energy. 1 mark
So there is a much greater frequency of successful collisions per unit time. 1 mark
The mark is for “many more particles have energy ≥ Ea” — the size of the effect matters. “Particles move faster” on its own scores nothing.
Draw the Maxwell–Boltzmann distribution curves for a fixed mass of a gas at two different temperatures. This gas decomposes when heated. By reference to these distribution curves, explain why the rate of decomposition of the gas increases at higher temperatures.
Show answer
The curves (as in the two-temperature figure above): label the vertical axis number of molecules and the horizontal axis (kinetic) energy. 1 mark Each curve starts at the origin and its tail approaches but never meets the energy axis. 1 mark
The higher-temperature curve has a lower peak shifted to the right, the same area beneath it, and crosses the first curve once. 1 mark
The explanation: at the higher temperature the molecules have more energy, so a greater proportion of molecules have energy ≥ the activation energy. 1 mark A greater proportion of collisions are therefore successful. 1 mark So the frequency of successful collisions increases and the rate increases. 1 mark
Do not accept a higher-temperature curve whose area is larger (the number of molecules is fixed), or one that touches the energy axis.
A student investigates the effect of temperature on the rate of reaction between sodium thiosulfate solution and dilute hydrochloric acid: Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + SO2(g) + S(s) + H2O(l). The student mixes the solutions in a flask placed on a paper marked with a cross, and records the time for the cross to disappear as the mixture becomes cloudy. The table shows the results.
| Temperature / °C | 22 | 31 | 36 | 42 | 49 | 54 |
|---|---|---|---|---|---|---|
| Time, t, for cross to disappear / s | 87 | 48 | 36 | 26 | 44 | 12 |
| 1/t / s−1 | 0.0115 | 0.0208 | 0.0278 | 0.0385 | 0.0227 |
The student uses a stopwatch to measure the time. The stopwatch shows each time to the nearest 0.01 s. Suggest why the student records the times to the nearest second and not to the nearest 0.01 s.
Show answer
It is hard to judge the exact moment the cross disappears (or: the mixture becomes too cloudy to judge that precisely / reaction time when stopping the clock). 1 mark
Vague appeals to “accuracy” are ignored unless qualified by the end-point judgement.
The rate of reaction is proportional to 1/t. Complete the table above.
Show answer
1/t = 1 ÷ 12 = 0.0833 s−1 1 mark
Plot the values of 1/t against temperature on the graph below. Draw a line of best fit.
Show answer
All six points plotted correctly (±½ small square). 1 mark
A smooth best-fit curve that misses the anomalous point at 49 °C and passes within one small square of the other five. 1 mark
The best-fit line is penalised if it is forced through (0, 0). Spot the anomaly, plot it, then ignore it.
Use your line of best fit to estimate the time for the cross to disappear at 40 °C. Show your working.
Show answer
from the line, 1/t at 40 °C ≈ 0.035 s−1, so t = 1 ÷ 0.035 ≈ 29 s 1 mark
Any answer consistent with your own line scores — but the working must show the 1/t read-off being inverted.
Suggest, by considering the products of this reaction, why small amounts of reactants are used in this experiment.
Show answer
SO2 is toxic — small amounts limit how much of the gas forms. 1 mark
“Harmful” or “causes acid rain” is not enough — the credited word is toxic/poisonous.
The student could do the experiment at lower temperatures using an ice bath. Suggest why the student chose not to carry out experiments at temperatures in the range 1–10 °C.
Show answer
The reaction would be very slow / take too long (and the gradual end-point becomes even harder to judge). 1 mark
Potassium manganate(VII) reacts with sodium ethanedioate in the presence of dilute sulfuric acid: 2MnO4−(aq) + 16H+(aq) + 5C2O42−(aq) → 2Mn2+(aq) + 8H2O(l) + 10CO2(g). The mixture is purple at the start and goes colourless when all the MnO4−(aq) ions have reacted, so the rate can be measured as 1000/t, where t is the time taken to go colourless. A student timed the reaction at different temperatures, with the same concentrations and volumes of each reagent every time.
| Temperature / °C | 32 | 38 | 44 | 54 | 67 |
|---|---|---|---|---|---|
| Time t / s | 155 | 85 | 50 | 22 | 9 |
| 1000/t / s−1 | 6.45 | 11.8 | 20.0 | 45.5 |
Complete the table above.
Show answer
1000 ÷ 9 = 111 s−1 (111.1) 1 mark
Any correctly rounded number of significant figures is allowed (110 is fine); a recurring-dot answer is not.
State the independent variable in this investigation.
Show answer
Temperature 1 mark
The student noticed that the temperature of each reaction mixture decreased during each experiment. Suggest how the student calculated the temperature values in the table above.
Show answer
Measure the temperature at the start and at the end (or at regular intervals) and take the mean. 1 mark
Use the data in the table to plot a graph of 1000/t against temperature.
Show answer
A vertical scale that uses more than half the axis for the five points. 1 mark
Points plotted correctly (±½ small square each). 1 mark
A smooth best-fit curve, rising throughout, within one small square of each point. 1 mark
Use your graph to find the time taken for the mixture to go colourless at 60 °C. Show your working.
Show answer
from the curve, 1000/t at 60 °C ≈ 68, so t = 1000 ÷ 68 ≈ 15 s 1 mark
Use the value from your line (construction lines shown on the graph); give the answer to at least 2 significant figures.
Apparatus is set up to measure the time for 20.0 cm3 of sodium thiosulfate solution to react with 5.0 cm3 of hydrochloric acid in a 100 cm3 conical flask at 20 °C. The timer is started when the thiosulfate is added and stopped when the cross can no longer be seen. What is likely to decrease the accuracy of the experiment? Tick (✓) one box.
The same experiment is repeated at 20 °C using a 250 cm3 conical flask instead of the 100 cm3 one. Which statement is correct about the time taken for the cross to disappear when using the larger flask? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Concentration & pressure
These two levers share one mechanism, and it is a different mechanism from temperature — which is exactly why they are worth grouping together.
Increasing the concentration of a solution puts more particles into the same volume. The particles are closer together, so they collide more frequently — and more collisions per second means a faster rate. Nothing about the energy of those collisions changes.
Increasing the pressure of a gas does the same thing by a different route: squeezing a fixed number of gas particles into a smaller volume packs them closer together. Higher pressure simply is higher concentration for a gas — so, same mechanism, more frequent collisions, faster rate.
That is the clean line to hold: concentration and pressure change collision frequency only. They do not touch the activation energy or the Maxwell–Boltzmann curve, so the fraction of successful collisions is unchanged — there are just more of them. Contrast that with temperature and catalysts, which change that fraction. (At GCSE you also met surface area of a solid; it belongs to the same frequency family — more exposed particles, more collisions — but A-Level frames the levers as concentration and pressure.)
To see the concentration effect, AQA's named route is continuous monitoring of calcium carbonate + hydrochloric acid — follow the mass lost as CO2 escapes, and repeat at different acid concentrations. A steeper initial gradient is the faster reaction.
Interactive — crowd the box, count the collisions
Ea and the fraction of collisions that succeed are untouched — there are simply more collisions each second.
- Frequency, not energy. Concentration and pressure change how often particles collide — do not mention the Maxwell–Boltzmann curve or activation energy in these explanations. That is the temperature/catalyst story, not this one.
- Pressure is gases only. Frame it as “more particles per unit volume” / “particles closer together” — the same statement as increasing concentration.
- More frequent ≠ more energetic. Crowding particles does not give them more energy; every collision is exactly as (un)likely to succeed as before.
GCSE groundwork: C6 — factors affecting rate (including surface area); concentration as mol dm−3 in Amount of Substance.
Looking ahead: raising the pressure or concentration speeds a reaction up here — but for a reversible reaction it also shifts the position of equilibrium, which is a separate idea from the rate.
🧪 Exam-style questions
An excess of magnesium reacts with hydrochloric acid to form hydrogen gas. Line X on the graph shows how the volume of hydrogen produced changes with time as magnesium reacts with 30 cm3 of 1.0 mol dm−3 hydrochloric acid. The reaction is repeated using 20 cm3 of 2.0 mol dm−3 hydrochloric acid, with all other conditions the same. Which line shows how the volume of hydrogen produced changes with time? Tick (✓) one box.
Magnesium reacts with an acid to form hydrogen gas. Line X on the graph shows how the volume of hydrogen gas varies with time when 50 cm3 of 0.50 mol dm−3 acid reacts with an excess of magnesium. The reaction is repeated under the same conditions but using 25 cm3 of 1.50 mol dm−3 acid. The magnesium is in excess. Which line represents this second reaction? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Catalysts
A catalyst speeds a reaction up and comes out the other side unchanged — and the how is a Maxwell–Boltzmann story, which is why it belongs at the end of this page.
Catalyst — a substance that increases the rate of a reaction without being changed in chemical composition or amount at the end of the reaction.
A catalyst takes part in the reaction and is then regenerated, so it is not used up — a tiny amount keeps working. And it changes only the speed: it does not change how much product forms, and it does not change the enthalpy change, ΔH.
The mechanism is one sentence: a catalyst provides an alternative reaction route with a lower activation energy. Lower the bar, and — reading straight off the Maxwell–Boltzmann curve — a greater proportion of particles now have energy ≥ Ea, so a greater proportion of collisions succeed, and the reaction goes faster. The catalyst does not push the curve or heat anything up; it moves the line.
“Explain, with reference to the Maxwell–Boltzmann distribution, how a catalyst increases the rate of reaction.”
A catalyst provides an alternative route (mechanism) with a lower activation energy. 1 mark So a greater proportion of particles have energy ≥ Ea. 1 mark A greater proportion of collisions are therefore successful, and the rate increases. 1 mark
Do not write “the catalyst gives the particles more energy” — it lowers the bar they must clear, it does not raise their energy.
Which substance actually catalyses which reaction — and the split into homogeneous and heterogeneous catalysis — is Year 2 detail. Here, the definition and the Maxwell–Boltzmann explanation are the whole requirement.
- Explain it through the Maxwell–Boltzmann distribution. The credited answer is “a greater proportion of particles have energy ≥ the (new, lower) Ea” — not “gives particles more energy”.
- Lower barrier by an alternative route. A catalyst does not lower the energy of the reactants and does not change ΔH — it offers a different path with a smaller Ea.
- Keep the definition complete. “Not used up” / “unchanged in chemical composition and amount” is part of what a catalyst is.
GCSE groundwork: C6 — catalysts.
🧪 Exam-style questions
The rate of a reaction is greater when a catalyst is used, without changing the temperature. Which statement explains why? Tick (✓) one box.
Source: AQA A-Level Chemistry past papers.
Capstone: one reaction, every lever
Every factor on this page speeds a reaction up — but not for the same reason, and the difference is the whole point. Line them up and each one is doing exactly one of two jobs: changing how often particles collide, or changing what fraction of those collisions succeed.
| Lever | What it changes | Why | What you’d see |
|---|---|---|---|
| Raise the temperature | Success fraction (mostly) + a little frequency | a greater proportion of particles have E ≥ Ea | steeper initial gradient; a small rise gives a large speed-up |
| Increase the concentration | Collision frequency | more particles per unit volume → collisions more often | steeper initial gradient |
| Increase the pressure (gases) | Collision frequency | same particles in a smaller volume = higher concentration | steeper initial gradient |
| Add a catalyst | Success fraction (lowers Ea) | an alternative route with a lower barrier → more particles clear it | faster; ΔH and final yield unchanged |
| Increase surface area (solids) | Collision frequency | more particles exposed at the surface to be hit | steeper initial gradient |
Frequency levers — concentration, pressure and surface area — add more collisions without changing the odds on any single one: the Maxwell–Boltzmann curve and the activation energy never move. Success-fraction levers — temperature and catalysts — change the odds themselves: temperature pushes more particles over a fixed Ea, and a catalyst lowers the Ea they must clear.
Temperature is the only lever that does a little of both — but it is dominated by the success-fraction effect, and that is the distinction the examiner reports say students blur. Sort any rate question into “more collisions” or “more successful collisions” first, and the right explanation — and the marks — follow.
Interactive — diagnose the lever
1 · which mechanism is at work?
2 · and the sentence that earns the mark?
- A collision reacts only with energy ≥ Ea and the correct orientation; most collisions fail the energy test.
- Activation energy (Ea) = the minimum energy colliding particles must have to react — the barrier height on the reaction profile.
- Rate of reaction = change in concentration of a reactant or product per unit time; units are always “something per time” (mol dm−3 s−1, cm3 s−1, g s−1…).
- Measuring rate: continuous monitoring (gas volume or mass loss) → read the gradient with a tangent, quoting units; or a clock reaction → time to a fixed change, rate ∝ 1/time.
- Maxwell–Boltzmann: starts at the origin (no particle has zero energy), peaks at the most probable energy, mean just to its right, tail approaches but never touches the axis (no maximum); area = number of molecules.
- The particles that can react are the area with E ≥ Ea — a small slice at ordinary temperatures.
- Temperature ↑: a greater proportion of particles have E ≥ Ea (dominant) plus slightly more frequent collisions; the reactive area grows sharply, so a small rise gives a large rate increase. The M–B area is conserved.
- RP3: sodium thiosulfate + HCl, time a cross to disappear at different temperatures; rate ∝ 1/time.
- Concentration / pressure ↑: more particles per unit volume → more frequent collisions only (Ea and the M–B curve are unchanged); for gases, more pressure = higher concentration.
- Catalyst: an alternative route of lower Ea → a greater proportion of particles clear it → faster; not used up, and ΔH is unchanged.
- The unifier: concentration, pressure and surface area change collision frequency; temperature and catalysts change the fraction of successful collisions.