What this chapter covers8.1, 8.2, 8.3
Two questions can be asked of any reaction, and they have nothing to do with each other. How far does it go? is answered by energetics and equilibrium. How fast does it get there? is kinetics, and it is the whole of topic 8. A reaction can be violently thermodynamically favourable and still sit untouched for a thousand years — a lump of coal in air is the standard example — because favourable says nothing about fast.
The topic is built on one idea and then applied three times. The idea is that molecules only react when they collide hard enough and in the right way, so the rate of a reaction is the rate at which useful collisions happen. Everything else follows: concentration and pressure change how often molecules meet, temperature changes how hard they hit, and a catalyst changes how hard they need to hit.
What topic 8 asks you to do
8.1 Rate of reaction
Explain and use the terms rate of reaction, frequency of collisions,
effective collisions and non-effective collisions; explain qualitatively, in
terms of the frequency of effective collisions, the effect of concentration and pressure changes on
the rate of a reaction; and use experimental data to calculate the rate of a reaction.
8.2 Effect of temperature on reaction rates and the concept of
activation energy
Define activation energy, EA, as the minimum energy required for a collision to
be effective; sketch and use the Boltzmann distribution to explain the significance of activation
energy; and explain qualitatively, in terms both of the Boltzmann distribution and of the frequency
of effective collisions, the effect of temperature change on the rate of a reaction.
8.3 Homogeneous and heterogeneous catalysts
Explain and use the terms catalyst and catalysis: explain that, in the presence of
a catalyst, a reaction has a different mechanism, one of lower activation energy; explain this
catalytic effect in terms of the Boltzmann distribution; and construct and interpret a reaction
pathway diagram for a reaction in the presence and absence of an effective catalyst.
Read those three statements again and notice what is missing. There are no rate equations, no orders of reaction, no rate constants and no half-lives. There is no Arrhenius equation. At AS the whole topic is qualitative except for one arithmetic skill — getting a rate out of a set of measurements — and the marks go to explanations written in the syllabus's own vocabulary.
What is deliberately not here
Rate equations, orders of reaction, rate constants, half-lives, the rate-determining step, the experimental determination of orders, and the mode of action of a heterogeneous catalyst in terms of adsorption all belong to A Level reaction kinetics, a separate unit later in the course. They are not in topic 8 and are not on this page. Where a model here does arithmetic on a curve or a distribution, it is doing work you are not asked to reproduce — the number is there so you can watch the qualitative claim come out true.
One sentence that earns most of the marks
Almost every explanation in this topic is a variation on: the rate depends on the frequency of effective collisions. A change either makes collisions more frequent, or makes a larger proportion of them effective, or both. Decide which of those two a given change does before you write anything, and the answer writes itself. Confusing the two — saying a temperature rise works because "the particles collide more often" and stopping there — is the single most common way to lose marks in topic 8.
What rate of reaction means8.1.1
Rate is a change divided by the time it took. In kinetics the change that matters is a change in amount of a substance, and the standard way of expressing it is per unit volume, so that the number does not depend on how big the flask is.
Rate of reaction
The rate of reaction is the change in the concentration of a reactant or a product divided by the time taken for that change.
rate = change in concentration ÷ time taken
Its units follow directly from that: concentration in mol dm⁻³ divided by time in s gives mol dm⁻³ s⁻¹. A reactant is being used up, so its concentration falls; the rate is quoted as a positive number and which species it refers to is stated.
Rate is not one number
A reaction does not proceed at a steady speed. It is fastest at the very beginning, when the reactants are at their highest concentration, and it slows continuously as they are consumed, reaching zero when one of them runs out. So "the rate of this reaction" is meaningless without a time attached to it. Three different quantities get called "the rate", and an exam question will tell you which one it wants:
| Quantity | What it is | How you get it |
|---|---|---|
| Initial rate | the rate at the instant of mixing, t = 0 | gradient of the tangent drawn at the origin |
| Rate at time t | the rate at one particular moment | gradient of the tangent drawn at that point |
| Average rate | the mean rate over an interval | total change ÷ total time — the gradient of the straight chord |
Initial rate is the one worth most
When two experiments are being compared — two concentrations, two temperatures, catalyst and no catalyst — the comparison is almost always made at t = 0. That is the only moment at which the two mixtures are identical except for the one thing you changed. Ten seconds later they differ in how much reactant is left as well, and the comparison is contaminated.
Frequency of collisions, and which ones count
The syllabus asks for four terms by name in outcome 8.1.1, and they are worth learning as a set because they are the vocabulary every later explanation is written in.
The four terms of 8.1.1
Rate of reaction — the change in concentration of a reactant or product per unit time.
Frequency of collisions — the number of collisions between particles per unit time.
Effective collision — a collision that results in a reaction: the colliding particles have at least the activation energy and are correctly oriented.
Non-effective collision — a collision that does not result in a reaction, because the particles had too little energy, or the wrong orientation, or both. They simply bounce apart unchanged.
The gap between the second and the third of those is enormous, and it is the reason chemistry exists at all. In a gas at room temperature an individual molecule collides with others of the order of 10⁹ times a second. If every collision reacted, every reaction would be over before you could put the stopper in. Almost all of them are non-effective.
Collision theory8.1.1
Collision theory is the model underneath the whole topic. It makes three claims, and every explanation in topic 8 is an appeal to one of them.
Collision theory
For a reaction to occur:
- the particles must collide;
- they must collide with at least the activation energy;
- they must collide with the correct orientation.
Most collisions do not lead to reaction, because they fail the second or third condition.
The first condition is nearly free — particles in a liquid or a gas are colliding constantly. The second is the expensive one, and it is what topic 8.2 is about. The third is a geometry problem: existing bonds have to be stretched and broken and new ones formed, and that can only happen if the reacting parts of the two particles are the parts that meet.
"They need to collide" is not an answer
Asked why a change increases the rate, a very common answer is "because the particles collide more". On its own that earns nothing, because it is true of almost every change and it does not distinguish the two mechanisms. Say which: more frequent collisions (concentration, pressure) or a greater proportion of collisions being effective (temperature, catalyst) — and for temperature, both, with the second dominating.
Why "at least" matters in the definition
The activation energy is a minimum. A collision carrying exactly EA is effective; so is one carrying ten times as much. The condition is a threshold, not a target, which is why the useful quantity later on is the proportion of particles with energy ≥ EA — an area under a curve, not a point on it.
Measuring a rate in the laboratory8.1.3
Concentration is rarely what you actually measure. You measure something that changes as the reaction proceeds and that you can follow without disturbing it, then convert. Choosing that something is the practical half of outcome 8.1.3, and it is decided entirely by what the reaction does.
| Method | What is measured | Use it when | Watch out for |
|---|---|---|---|
| Gas collection | volume of gas, with a syringe or over water | a gas is produced | gas dissolving if collected over water — CO₂ and SO₂ both do |
| Loss of mass | mass of the flask on a balance | a gas is produced and it is heavy enough to register | useless for hydrogen — 2 g mol⁻¹ barely moves the balance |
| Colorimetry | absorbance of light | one species is coloured and the others are not | needs a calibration curve to turn absorbance into concentration |
| Titration of samples | concentration, directly | nothing observable changes | the sample must be quenched, or it keeps reacting in the flask |
| Clock reaction | the time to reach a fixed visible point | a precipitate or a sudden colour appears | gives 1/t as a measure of rate, not the rate itself |
| pH meter | pH, hence [H⁺] | H⁺ is made or used up | pH is logarithmic — convert before treating it as concentration |
Quenching
If you pipette a sample out of a reacting mixture and walk it to a burette, the reaction carries on inside the conical flask while you titrate, and the answer you get is for some later, unknown time. The sample has to be stopped first — run it into a large volume of ice-cold water to dilute and cool it, or add a reagent that removes one of the reactants. A question that gives you a titration method and asks for "one necessary precaution" is asking for this.
Clock reactions, and what 1/t really is
In a clock reaction you time how long the mixture takes to reach one fixed, easily seen point — a cross drawn on paper disappearing under a sulfur precipitate, or a sudden blue-black from starch and iodine. Up to that point the same amount of product has always been formed, whatever the conditions. So the average rate over that interval is a fixed quantity divided by t, which means 1/t is proportional to the average rate. Plotting 1/t against concentration or temperature is a legitimate rate measurement, and it is much easier than drawing tangents.
The thiosulfate clock
Na₂S₂O₃(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO₂(g) + H₂O(l)
Sulfur is formed as a fine pale suspension. A flask of the mixture is stood over a pencil cross and the time is taken for the cross to become invisible. The cross disappears when a fixed quantity of sulfur has formed — the same quantity every time, because it is fixed by how much sulfur it takes to block that much light. So a run that takes half as long had, on average, twice the rate.
It is a good reaction to meet early because both of the 8.1 variables can be tested on it: dilute the thiosulfate and the time lengthens; warm the mixture and it shortens sharply.
Calculating a rate from data8.1.3
This is the one calculation topic 8 asks for. You are given a set of measurements, or a graph drawn from them, and asked for a rate at some point. The procedure is always the same.
The procedure
- Plot the measured quantity against time, and draw a smooth curve — not a zig-zag from point to point.
- For an average rate over an interval: divide the total change by the total time.
- For the rate at an instant: draw a tangent to the curve at that point, take a large triangle on it, and find its gradient.
- Convert the result into the units the question wants — usually mol dm⁻³ s⁻¹.
Draw the triangle large
The gradient of a tangent is the least accurate reading on a rate paper, because a small error in where you put the ruler becomes a large error in the answer. Extend the tangent right across the grid and take your Δy and Δx from as big a triangle as fits. Mark the triangle on the graph — examiners give a mark for showing it, and they cannot give it for a number that appears from nowhere.
Getting to mol dm⁻³ s⁻¹
A gas syringe gives cm³ s⁻¹, and a balance gives g s⁻¹. Neither is a rate of reaction until it has been converted to an amount, and then to a concentration in the reacting mixture.
Worked example — from a gas syringe to a rate
Magnesium is added to 50.0 cm³ of hydrochloric acid and the hydrogen collected. The tangent at t = 0 has a gradient of 1.8 cm³ s⁻¹. Room conditions; the molar gas volume is 24.0 dm³ mol⁻¹ (24 000 cm³ mol⁻¹). Give the initial rate with respect to HCl.
1 — volume of gas per second to moles of gas per second
n(H₂) per second = 1.8 ÷ 24 000 = 7.5 × 10⁻⁵ mol s⁻¹
2 — moles of gas to moles of acid, using Mg + 2HCl → MgCl₂ + H₂: one H₂ for every two HCl.
n(HCl) per second = 2 × 7.5 × 10⁻⁵ = 1.5 × 10⁻⁴ mol s⁻¹
3 — moles per second to concentration per second, dividing by the volume of the solution, 50.0 cm³ = 0.0500 dm³.
rate = 1.5 × 10⁻⁴ ÷ 0.0500 = 3.0 × 10⁻³ mol dm⁻³ s⁻¹
Quoted as: the initial rate of reaction is 3.0 × 10⁻³ mol dm⁻³ s⁻¹ with respect to HCl.
Divide by the volume of the solution, not the volume of the gas
Step 3 catches people every year. The concentration that is changing is a concentration in the 50.0 cm³ of acid; the 24 000 cm³ is a molar gas volume and has already done its job in step 1. Dividing by the wrong volume gives an answer that is out by a factor of several hundred, and it looks perfectly plausible.
Rate with respect to what?
The same experiment has different numerical rates depending on which substance you quote it for, and they are in the ratio of the stoichiometric coefficients. In the example above, hydrogen is appearing at 1.5 × 10⁻³ mol dm⁻³ s⁻¹ while HCl is disappearing at twice that. Neither is more correct; the question tells you which it wants, and an answer with no species named is incomplete.
Concentration8.1.2
Concentration is the easiest of the rate factors to explain, because it acts on the first of collision theory's three conditions and leaves the other two alone.
The explanation, in the syllabus's own terms
Increasing the concentration of a reactant puts more particles in the same volume. The particles are therefore closer together, so they collide more frequently. The proportion of collisions that are effective is unchanged — the particles have not been given any more energy — but since collisions happen more often, effective collisions also happen more often, and the rate increases.
That last step is the one to be careful about. Concentration does not make collisions better. It makes them more numerous, and a fixed fraction of a larger number is a larger number.
Why the effect is more than proportional
For a collision between two different reactants, both concentrations matter: double the concentration of A and collisions with B double; double B as well and they double again. Doubling everything — halving the volume of the whole mixture, say — therefore quadruples the collision frequency, not doubles it. That is what the model above counts, and it is worth watching once, because it is the intuition behind the second-order behaviour you meet at A Level. At AS you are only asked to say that the rate increases, and why.
Reading two runs on one set of axes
Comparing two concentrations on one graph is a standard question, and there are two separate things to say about it.
| Feature of the curve | What it tells you |
|---|---|
| the gradient at t = 0 | the initial rate — steeper means faster |
| the height the curve levels off at | the total amount of product, fixed by how much limiting reactant there was |
| the time taken to level off | how long the reaction took overall |
Same amount, or a different amount?
If the more concentrated run used the same number of moles in a smaller volume, both curves finish at the same height — same amount of product, reached sooner. If it used more moles, the concentrated run also finishes higher. Exam questions use both, and they are distinguished only by the quantities in the stem. Read it before you decide where your second curve ends.
Pressure8.1.2
Pressure appears in outcome 8.1.2 alongside concentration, and the reason they are in the same sentence is that for a gas they are the same variable wearing different clothes.
Pressure is concentration, for a gas
Compressing a fixed amount of gas into a smaller volume does not change how many molecules there are; it changes how many there are per unit volume. That is precisely what concentration means. So increasing the pressure of a gas-phase reaction increases the concentration of the gaseous reactants, the molecules are closer together, they collide more frequently, and the rate increases.
Written as an equation, pV = nRT rearranges to n/V = p/RT, and n/V is the concentration. At constant temperature, concentration is directly proportional to pressure. You are not required to use the ideal gas equation here — it belongs to topic 4 — but it is worth seeing once, because it turns "pressure is like concentration" from an assertion into an identity.
Pressure does not mean harder collisions
The wrong answer, and it feels right: "at higher pressure the particles are pushed together more forcefully, so they collide with more energy." They do not. At constant temperature the average kinetic energy of the molecules is exactly the same at 10 atm as at 1 atm — temperature is the measure of that energy. Compressing a gas changes how often molecules meet, not how hard. Anything else is a temperature effect, and pressure is not temperature.
Pressure only acts on gases
Putting a reaction between two solutions under pressure does essentially nothing, because liquids are almost incompressible: the concentrations do not change. If a question gives you a reaction in aqueous solution and asks about the effect of increasing the pressure, the answer is "little or no effect", and the reason is that the concentrations are unaltered. Only a reaction with at least one gaseous reactant responds.
Rate and yield are different questions — again
Raising the pressure on N₂(g) + 3H₂(g) ⇌ 2NH₃(g) does two unrelated things. It increases the rate, because the molecules are closer together and collide more often — that is this topic. It also shifts the position of equilibrium towards the side with fewer gas molecules, increasing the yield — that is topic 7. A question about the conditions used in an industrial process usually wants both, kept apart and named.
Activation energy8.2.1
Every reaction starts by breaking something. Even one that gives out heat overall has to put energy in first, to stretch and break the bonds in the reactants before the new, stronger bonds can form and pay it back. That up-front cost is the activation energy, and it is the reason a mixture of petrol and air can sit in a tank indefinitely.
Activation energy
Activation energy, EA, is the minimum energy required for a collision to be effective.
That is the syllabus's wording for outcome 8.2.1 and it is the wording to reproduce. It is quoted in kJ mol⁻¹ and it is a property of the reaction, not of the conditions: warming a mixture does not lower its activation energy, it only supplies more particles that can meet it.
Three wordings that lose the mark
"The energy needed to start a reaction." Too vague — it sounds like something you supply
once with a match.
"The energy given out when particles collide." Backwards; the activation energy is put in,
not given out.
"The average energy of the particles." A different quantity altogether — and one that most
reacting particles are well above or well below.
Write minimum, write collision, write effective. Those three words are
the mark scheme.
Why a barrier exists at all
Follow two particles through a collision that works. They approach; the existing bonds begin to stretch and weaken, which costs energy, and the new bonds have not yet formed to repay it. At the worst moment the system is at its highest energy — old bonds half broken, new bonds half made — and if the collision brought in enough kinetic energy to reach that point, the arrangement can tip forward into products and release the balance. If it did not, the stretched bonds simply spring back and the particles separate unchanged. That is a non-effective collision, described from the inside.
Activation energy and enthalpy change are independent
EA is the height of the barrier. ΔH is the difference in level between the two sides. Knowing one tells you nothing about the other: there are exothermic reactions with enormous barriers (the combustion of coal) and endothermic ones with small barriers (ammonium nitrate dissolving). A question that gives you ΔH and asks you to deduce the rate is testing whether you know that.
Reaction pathway diagrams8.2.1, 8.3.1(c)
A reaction pathway diagram — also called an energy profile — plots the energy of the reacting system against the progress of the reaction. It is the standard way of showing an activation energy, and outcome 8.3.1(c) asks you to construct and interpret one.
What has to be on the diagram
- Axes: energy on the vertical, progress of reaction (or reaction pathway) on the horizontal. The horizontal axis is not time and has no scale.
- Two levels, one for reactants and one for products, each labelled.
- A hump between them, with its peak above both levels.
- EA, an arrow from the reactant level up to the peak.
- ΔH, an arrow from the reactant level to the product level, pointing down for exothermic and up for endothermic.
Both arrows start from the reactants
The commonest drawing error is an EA arrow drawn from the bottom of the diagram, or from the product level, up to the peak. It starts at the reactant level, because that is the energy the colliding particles already have; the activation energy is the extra they need. The ΔH arrow starts there too. Get both feet on the reactant line and the diagram is usually right.
The reverse reaction comes free
Read the same diagram from right to left and you have the reverse reaction. Its activation energy is the height from the product level to the same peak, so
EA(reverse) = EA(forward) − ΔH
For an exothermic forward reaction ΔH is negative, so the reverse barrier is the larger of the two — which is exactly why exothermic reactions are hard to reverse. The model above computes this both ways; try setting ΔH positive and watch which barrier wins.
Particles do not share energy equally8.2.2
Temperature is a measure of the average kinetic energy of the particles in a sample. The word average is doing a great deal of work. At any instant the particles in a flask of gas or solution have wildly different energies: some are nearly stationary, most are somewhere near the middle, and a few are travelling several times faster than the average. Collisions redistribute the energy constantly, so an individual particle is slow one moment and fast the next — but the distribution across the whole sample stays the same as long as the temperature does.
Why this is the whole of 8.2
If every particle had the same energy, a reaction would be all-or-nothing: below a threshold temperature, nothing at all; one degree above it, instantaneous. Reactions are not like that, and the reason is the spread. At any temperature there is a minority of particles at the fast end of the distribution, and those are the ones that react. Raising the temperature does not lift everyone over the barrier — it enlarges that minority, and it enlarges it much faster than it raises the average.
The Boltzmann distribution8.2.2
Plot the number of particles against energy and you get the curve that the whole of this topic turns on. The syllabus calls it the Boltzmann distribution; you will also see it called the Maxwell–Boltzmann distribution, and the two names mean the same curve here.
What each feature means
| Feature | What it is | Why |
|---|---|---|
| It starts at the origin | no particles have zero energy | the particles are all moving; a stationary particle is struck by its neighbours within nanoseconds |
| The peak | the most probable energy | more particles have this energy than any other — it is not the average |
| The mean | the average energy, to the right of the peak | the long tail pulls the mean above the mode, as it does in any skewed distribution |
| The long tail | a small number of very energetic particles | energy is shared by chance in collisions, and a few particles get an unusually large share |
| It never touches the axis | the curve is asymptotic to it | there is no maximum energy — however high you look, a vanishingly small number are up there |
| The total area | the total number of particles | every particle has some energy, so every particle is counted somewhere under the curve |
| The area beyond EA | the number able to react | this is the quantity the whole topic is about |
The sentence to carry into the exam
The shaded area to the right of EA represents the number of particles with energy equal to or greater than the activation energy — the particles whose collisions can be effective. Anything that increases that area increases the rate.
Energy on the x-axis, not speed
A distribution of molecular speeds and a distribution of molecular energies are different curves with different shapes, and the syllabus asks for energy. The tell-tale is the left-hand side: an energy distribution rises very steeply from the origin, while a speed distribution leaves the origin gently, curving away like the start of a parabola. Label the horizontal axis energy — or, if you want to be exact, kinetic energy — and the vertical axis number of particles. Neither axis gets a numerical scale.
Sketching the distribution8.2.2
Outcome 8.2.2 says sketch and use. The sketch is worth marks on its own, and they are awarded against a short list of features that is very nearly the same every year. Work through the list below on your own drawing before you decide it is finished.
When you are asked to add a second curve
A second curve, for a higher temperature, must satisfy three things at once: the peak is lower, the peak is further right, and the area under it is the same as under the first. The third is what most sketches get wrong — a higher-temperature curve drawn with the same peak height has more molecules in it than the flask contains. Draw it flatter and broader, crossing the first curve once on each side of the peak, and the areas look right. Label which curve is which.
Why the peak has to drop
The total number of particles is fixed, and the total area is that number. Heating spreads the particles over a wider range of energies. The same area spread over a wider base must be lower — there is no other way for both statements to be true. That is the whole argument, and it is worth being able to give it in one sentence, because "why does the peak get lower?" is a common follow-up.
Temperature and the distribution8.2.3
Raise the temperature and the whole distribution changes shape. This is the part of topic 8 with the most marks attached to it, and the answer has two halves that have to be given in the right proportion.
The two effects of a temperature rise, in order of importance
1 — More particles have energy ≥ EA. The distribution flattens and shifts to the right, so the area beyond EA grows sharply. A much larger proportion of collisions is therefore effective. This is the major effect.
2 — The particles move faster, so they collide more frequently. More collisions per second means more chances to react. This is the minor effect.
Both increase the rate, and both should be mentioned — but a full-mark answer says which one dominates.
The half-answer that keeps losing marks
"At a higher temperature the particles move faster and collide more often, so the rate increases." Every word of that is true, and it describes the effect that accounts for only a few per cent of what actually happens. If the answer stops there it has missed the point of the section. The energy argument comes first and the collision-frequency argument is the footnote — the next section shows the arithmetic that justifies saying so.
Do not say the activation energy falls
Heating a mixture does not lower EA. The barrier is fixed by the bonds being broken and made; the temperature decides how many particles can get over it. On a sketch, the vertical EA line stays exactly where it was and the curve moves underneath it. Only a catalyst moves that line, and it does so by providing a different route, not by lowering the barrier of the original one.
Why 10 K nearly doubles the rate8.2.3
The rule of thumb is that a rise of about 10 K roughly doubles the rate of a reaction near room temperature. That is a striking claim: 298 K to 308 K is an increase in absolute temperature of about 3%, and the average kinetic energy goes up by the same 3%. Something has to explain how a 3% change produces a 100% change.
Where the factor of two comes from
Not from the average. From the tail. The proportion of particles beyond EA depends on the temperature exponentially, so a small shift in the curve moves a disproportionate number of particles across the line. At an activation energy of around 50 kJ mol⁻¹ — an entirely ordinary value — the model above puts the increase at roughly 90% for ten degrees, while the collision frequency, which goes as √T, rises by under 2%. The approximately fiftyfold difference between those two numbers is the reason the energy effect is called the major one.
The rule is a rule of thumb, not a law
"Doubles" is an approximation that holds near room temperature for activation energies in the region of 50 kJ mol⁻¹. Change either and the factor changes: a reaction with a large activation energy is far more temperature-sensitive than one with a small one, which is exactly why the model above lets you vary EA. You are not asked to calculate any of this at AS, and you should not quote "the rate doubles" as though it were exact. What you are asked for is the explanation — a small temperature rise produces a large increase in the number of particles with energy ≥ EA.
A full-mark answer, assembled
"Explain why increasing the temperature from 25 °C to 35 °C approximately doubles the rate of this reaction. [3]"
At the higher temperature the particles have a greater average kinetic energy, so the Boltzmann distribution is flatter and shifted to the right. (1) A significantly larger proportion of the particles therefore has energy equal to or greater than the activation energy, so a much larger proportion of collisions is effective. (2) The particles are also moving faster and so collide more frequently, but this is a minor effect compared with the increase in the number of particles able to react. (3)
Three sentences, three marks, and the third one is where the mark for "which effect dominates" lives. Note what is not in it: no claim that EA changed, and no numbers.
What a catalyst does8.3.1
A catalyst speeds a reaction up and is still there, chemically unchanged, when the reaction has finished. That second half is the surprising one, and it is what separates a catalyst from a reactant.
Catalyst and catalysis
A catalyst is a substance that increases the rate of a chemical reaction without itself being chemically changed at the end of the reaction. It provides an alternative reaction pathway, with a lower activation energy.
Catalysis is the process of increasing the rate of a reaction by using a catalyst.
Note at the end. A catalyst may well be changed during the reaction; it has to be regenerated by a later step, and often is not present in its original form while the reaction is running. What matters is the balance sheet at the finish.
A catalyst does not change the yield
Both curves in that comparison level off at the same height. A catalyst changes how fast equilibrium or completion is reached, not where it is: it lowers the barrier for the forward and the reverse reaction by the same amount, so the position of equilibrium is untouched and the enthalpy change is untouched. "Adding a catalyst increases the yield" is wrong, and it is wrong in topic 7 as well as here.
Three things a catalyst leaves alone
The activation energy of the original reaction — that barrier is still there; the catalyst has supplied a different route past it. The Boltzmann distribution — the particles have not been given any more energy, so the curve does not move. The enthalpy change — the reactant and product levels on a pathway diagram are exactly where they were. Each of those three is a favourite short question.
An alternative mechanism8.3.1(a)
The wording of outcome 8.3.1(a) is precise and it is worth matching: in the presence of a catalyst, a reaction has a different mechanism — one of lower activation energy.
A mechanism is the actual sequence of steps by which a reaction happens. The uncatalysed reaction has one mechanism with one barrier. The catalyst opens a second, quite different sequence of steps, in which the catalyst itself takes part: it is consumed in one step and regenerated in a later one. Every step of that new sequence has a lower barrier than the original single step, so far more collisions have enough energy to get through it.
The Contact process — a mechanism you can write down
Sulfur dioxide is oxidised to sulfur trioxide over vanadium(V) oxide:
SO₂(g) + ½O₂(g) → SO₃(g)
The catalysed route is two steps. The catalyst oxidises the sulfur dioxide and is itself reduced:
SO₂(g) + V₂O₅(s) → SO₃(g) + V₂O₄(s)
and is then re-formed by oxygen:
V₂O₄(s) + ½O₂(g) → V₂O₅(s)
Add the two steps together and the vanadium oxides cancel, leaving the overall equation unchanged. That cancellation is the test of a catalytic mechanism: the catalyst appears on the left of one step and the right of another, so it is absent from the sum.
How to check a proposed mechanism in a question
Add the steps. The catalyst must cancel completely, and what is left must be the overall equation with nothing missing. If the catalyst survives on one side of the sum it is being consumed and is a reactant, not a catalyst. If the sum does not reproduce the overall equation, a step is wrong.
The catalysed pathway diagram8.3.1(c)
Outcome 8.3.1(c) asks for a reaction pathway diagram drawn for a reaction in the presence and absence of an effective catalyst. Both routes go on one set of axes.
What the two-route diagram must show
- the same reactant level and the same product level for both routes — the catalyst does not change ΔH;
- the uncatalysed route as one hump, the taller one;
- the catalysed route as a lower hump, dashed or otherwise distinguished, and labelled;
- both activation energies marked from the reactant level to their own peaks, labelled EA and EA(catalysed).
Two humps, not one hump lowered
A catalysed route that happens in two steps is drawn with two peaks and a dip between them, the dip being the intermediate that forms and is then used up. Both peaks are below the uncatalysed one. A single lowered hump is accepted where the question does not specify a two-step mechanism, but if the question gives you two steps, draw two peaks — and remember the dip is an intermediate, not the product.
The product level does not move
Drawing the catalysed curve finishing lower than the uncatalysed one is a common slip, and it says that the catalyst changed the enthalpy of the products — which would mean it changed what the products are. Same reactants, same products, same two levels; only the route between them differs.
Catalysts and the Boltzmann distribution8.3.1(b)
Outcome 8.3.1(b) asks for the catalytic effect explained in terms of the Boltzmann distribution, and this is the one explanation in topic 8 where the curve stays still and the line moves.
The explanation
The catalyst provides an alternative route with a lower activation energy. On the Boltzmann distribution the EA line therefore moves to the left. The curve itself is unchanged — the particles have the same energies as before, because the temperature has not changed. But a much larger area now lies to the right of the line, so a much greater proportion of the particles has enough energy to react, a greater proportion of collisions is effective, and the rate increases.
Temperature moves the curve; a catalyst moves the line
These two explanations use the same diagram and are easy to swap by mistake.
Temperature up: the curve flattens and shifts right; the EA
line does not move.
Catalyst added: the line moves left; the curve does not move.
Both end with a larger shaded area, and the shaded area is the point in both cases. Say which one moved.
Why the effect is so large
The same exponential steepness that makes ten degrees worth a factor of two makes a modest cut in the activation energy worth a factor of thousands. Dropping EA from 50 to 30 kJ mol⁻¹ at room temperature multiplies the reacting fraction by about 2500 — and the cut costs nothing, because the catalyst is not consumed. That is why an industrial catalyst can let a plant run at 450 °C instead of at a temperature no material would survive. The model above integrates both tails, so you can push the numbers around and watch the ratio.
Heterogeneous catalysts8.3.1
Heterogeneous catalyst
A heterogeneous catalyst is in a different phase from the reactants. In practice this almost always means a solid catalyst with gaseous or aqueous reactants, so the reaction happens at the surface.
Phase, not state, is the word the definition turns on — two immiscible liquids are two phases even though both are liquids. In the cases you will meet, though, the test is simple: look at the state symbols, and if the catalyst's differs from the reactants', it is heterogeneous.
| Process | Catalyst | Phases |
|---|---|---|
| Haber process, N₂ + 3H₂ ⇌ 2NH₃ | iron | solid catalyst, gaseous reactants |
| Contact process, 2SO₂ + O₂ ⇌ 2SO₃ | vanadium(V) oxide | solid catalyst, gaseous reactants |
| Catalytic converter, 2CO + 2NO → 2CO₂ + N₂ | platinum, palladium, rhodium | solid catalyst, gaseous reactants |
| Hydrogenation of alkenes | nickel | solid catalyst, gaseous or liquid reactants |
| Poly(ethene) manufacture | Ziegler–Natta catalyst | solid catalyst, gaseous reactant |
How far the AS syllabus goes
At AS you need the definition, the phase comparison and examples. The mode of action of a heterogeneous catalyst — reactants adsorbing onto the surface at active sites, bonds weakening, the products desorbing — belongs to A Level reaction kinetics and is not examined in topic 8. It is worth one sentence of awareness and no more: the surface is where the alternative mechanism lives, which is why surface area and poisoning matter to an industrial catalyst.
Homogeneous catalysts8.3.1
Homogeneous catalyst
A homogeneous catalyst is in the same phase as the reactants — everything in one gaseous mixture, or everything dissolved in one solution.
Because catalyst and reactants are mixed at the molecular level, homogeneous catalysis is where the two-step mechanism is easiest to write out in full, and where exam questions usually ask for one.
Ozone destruction by chlorine radicals — all in the gas phase
Chlorine radicals from CFCs catalyse the breakdown of ozone in the stratosphere:
Cl• + O₃ → ClO• + O₂
ClO• + O → Cl• + O₂
Overall: O₃ + O → 2O₂. The chlorine radical is regenerated by the second step, so one radical destroys ozone molecules over and over — which is why a small quantity of CFC does so much damage. Catalyst and reactants are all gases: homogeneous.
Iron ions and the persulfate–iodide reaction — all in solution
The reaction between peroxodisulfate and iodide ions is very slow, because both ions are negatively charged and repel each other:
S₂O₈²⁻(aq) + 2I⁻(aq) → 2SO₄²⁻(aq) + I₂(aq)
Fe²⁺(aq) catalyses it by splitting the job into two steps, each between oppositely charged ions:
2Fe²⁺(aq) + S₂O₈²⁻(aq) → 2Fe³⁺(aq) + 2SO₄²⁻(aq)
2Fe³⁺(aq) + 2I⁻(aq) → 2Fe²⁺(aq) + I₂(aq)
Everything is aqueous, so the catalyst is homogeneous. Notice that the iron is genuinely changed during the reaction — oxidised in the first step, reduced back in the second. It ends as it started, which is all the definition requires.
Decide from the phases, not from the chemistry
The classification is a comparison of state symbols and nothing else. Vanadium(V) oxide is heterogeneous even though its mechanism is written out as two neat equations, because it is a solid in a gaseous mixture. Concentrated sulfuric acid catalysing an esterification is homogeneous because everything in the flask is liquid. Write the state symbols down before you answer.
Why industry cares8.3.1
The whole point of a catalyst on an industrial scale is that it buys the same rate at a lower temperature and, sometimes, a lower pressure. Everything follows from that.
| Consequence | Why it follows |
|---|---|
| lower energy demand | less fuel burnt to heat and compress the reactants |
| less carbon dioxide released | the energy usually comes from burning a fossil fuel |
| lower production costs | fuel is a large part of the running cost of a plant |
| cheaper plant | lower temperatures and pressures need less demanding materials |
| a better yield, indirectly | for an exothermic reaction a lower temperature also shifts the equilibrium towards the products — the catalyst makes that lower temperature affordable |
The indirect yield argument is worth understanding
A catalyst does not change the position of equilibrium. But in the Contact process the forward reaction is exothermic, so a lower temperature would give a better yield — and without a catalyst a low temperature would make the reaction uselessly slow. The catalyst does not raise the yield; it makes the temperature that raises the yield practical. Getting that distinction right is the difference between a good answer and a wrong one.
Enzymes
Enzymes are biological catalysts — proteins, highly specific to one reaction, working at body temperature and in a narrow pH range. They appear in the deck-era syllabus and in biology, and they are a fair illustration of catalysis in the real world, but the 2025–2027 outcomes for topic 8 do not ask for them. Know the word; do not spend revision time on active sites for this paper.
The four standard answers8.1, 8.2, 8.3
Nearly every explanation question in topic 8 is one of four, and they differ in exactly one respect: whether the change makes collisions more frequent, makes a larger proportion of them effective, or both. Learn the four side by side and the distinction stops being slippery.
| Change | Frequency of collisions | Proportion effective | What to write |
|---|---|---|---|
| Concentration ↑ | increases | unchanged | more particles per unit volume, so they are closer together and collide more frequently; the same proportion of a larger number of collisions is effective |
| Pressure ↑ (gases) | increases | unchanged | the same number of molecules in a smaller volume, so the concentration rises and they collide more frequently |
| Temperature ↑ | increases slightly | increases greatly | the distribution flattens and shifts right, so many more particles have energy ≥ EA and a much greater proportion of collisions is effective; they also collide more frequently, but that is the minor effect |
| Catalyst added | unchanged | increases greatly | an alternative mechanism of lower activation energy, so more of the existing particles have enough energy; the distribution itself does not move |
Two phrases to have ready
"the frequency of effective collisions" — the syllabus's own phrase, and the safest way to
finish any of these answers.
"energy equal to or greater than the activation energy" — not "more than", because
EA is a minimum and a particle with exactly that energy counts.
What never changes
Across all four rows: the activation energy of the original reaction, the enthalpy change, and the position of equilibrium. The first is moved by nothing on this page — a catalyst supplies a different route, it does not lower the original barrier. The second and third are moved only by temperature, and that belongs to topics 5 and 7.
Self-test8.1, 8.2, 8.3
Thirty questions across the whole of topic 8. Each one explains itself after you answer, and the explanation is usually the part worth reading — several of the questions are built around the traps in the sections above.
Definitions to learn8.1, 8.2, 8.3
| Term | Definition |
|---|---|
| Rate of reaction | the change in the concentration of a reactant or product per unit time; units mol dm⁻³ s⁻¹ |
| Frequency of collisions | the number of collisions between particles per unit time |
| Effective collision | a collision that results in reaction: the particles have at least the activation energy and collide with the correct orientation |
| Non-effective collision | a collision that does not result in reaction, because the particles had less than the activation energy or the wrong orientation |
| Activation energy, EA | the minimum energy required for a collision to be effective |
| Boltzmann distribution | the distribution of the kinetic energies of the particles in a sample at a given temperature |
| Catalyst | a substance that increases the rate of a reaction without itself being chemically changed at the end of the reaction, by providing an alternative pathway of lower activation energy |
| Catalysis | the process of increasing the rate of a reaction by using a catalyst |
| Homogeneous catalyst | a catalyst in the same phase as the reactants |
| Heterogeneous catalyst | a catalyst in a different phase from the reactants |
| Mechanism | the sequence of steps by which a reaction actually takes place |
| Reaction pathway diagram | a plot of the energy of the reacting system against the progress of the reaction, showing the activation energy and the enthalpy change |
Data used on this page8.1, 8.2
The models on this page compute rather than store their answers, and these are the constants and figures they compute from.
| Quantity | Value used | Where it is used |
|---|---|---|
| Gas constant, R | 8.31 J K⁻¹ mol⁻¹ | every fraction-above-EA calculation |
| Molar gas volume at room conditions | 24.0 dm³ mol⁻¹ | converting a gas volume to an amount |
| Standard comparison temperature | 298 K (25 °C) | the Boltzmann and ten-degree models |
| Typical activation energy | 50 kJ mol⁻¹ | the default in the ten-degree model, chosen because it is where the rule of thumb works |
| Collision frequency with temperature | proportional to √T | the minor effect in the ten-degree model |
These are reference values, not the official data booklet
Use the data booklet issued with your paper for any value in an examination. The figures here are consistent with one another so that the models give sensible answers, and they are close to the booklet's, but they are not a substitute for it. Activation energies in particular vary between sources for the same reaction, because they depend on the temperature range over which they were measured.
A note on the arithmetic in the models
The fraction of particles with energy ≥ EA is obtained here by integrating the Boltzmann energy distribution numerically, and the ratio of two such fractions is what the ten-degree model reports. That is A Level work dressed up as a demonstration: at AS you are asked for the qualitative statement — a small temperature rise gives a large increase in the number of particles able to react — and never for the number. The arithmetic is on the page so that the qualitative statement can be watched coming true, and so that "the major effect" can be seen to be major.