Some reactions are over before you can blink — an explosion, or a precipitate appearing the instant two solutions meet. Others, like the rusting of iron, take years. C6 asks two questions about every reaction: how fast it goes (the rate), and how far it goes (the extent — because some reactions can run backwards too). Industry asks exactly the same two questions of every process it runs, which is why this topic is full of real chemical engineering.
Within C6 itself (4.6) there is no Triple-only content — Combined Science and Triple Chemistry students learn the same material here, and the only split is by tier:
- Everyone — calculating mean rates, drawing and interpreting rate graphs, drawing tangents, the factors affecting rate, required practical 5, collision theory, catalysts, reversible reactions and dynamic equilibrium.
- Higher Tier H — calculating the gradient of a tangent as the rate at a specific time, using mol/s as a unit of rate, and the whole of section 7: Le Chatelier’s principle and the effects of changing concentration, temperature and pressure on equilibrium.
1Calculating Rates of Reaction
The rate of a reaction is how quickly reactants are turned into products. To measure it you track a quantity you can actually see changing — the mass of the reaction mixture (in g) or the volume of gas given off (in cm³) — and divide by the time taken.
mean rate of reaction = quantity of reactant usedtime taken
mean rate of reaction = quantity of product formedtime taken
The quantity can be a mass in g or a volume in cm³, so the units of rate are g/s or cm³/s. Higher Higher Tier students also use the quantity in moles, giving the unit mol/s.
Worked example — mean rate
Magnesium ribbon is added to dilute hydrochloric acid. The reaction produces 48 cm³ of hydrogen gas in the first 60 seconds.
- Write the equation: mean rate = quantity of product formedtime taken
- Substitute with units in place: mean rate = 48 cm³60 s
- Answer: 0.8 cm³/s. The unit does the checking for you — a volume divided by a time must come out in cm³/s.
Worked example — mean rate between two times
Most exam questions don’t start the clock at zero — they ask for the mean rate between two times read off a graph, often with a unit to convert. Reading a volume–time graph, the gas collected is 60 cm³ at 2 minutes and 84 cm³ at 4 minutes. Calculate the mean rate of reaction between these times, in cm³/s.
How the data usually appears in an exam: two readings taken off a volume–time curve. The mean rate between them is the change in volume (24 cm³) divided by the time (120 s).
- Change in quantity = 84 − 60 = 24 cm³ of gas formed in this interval.
- Time taken = 4 − 2 = 2 minutes. The answer is wanted per second, so convert first: 2 × 60 = 120 s.
- mean rate = 24 cm³120 s = 0.20 cm³/s (2 s.f.).
- Forget the minute–second conversion and you’d get 12 cm³/min — right arithmetic, wrong unit, no marks.
Examiners regularly see correct arithmetic lose a mark to a wrong or missing unit. Carry the unit through every line of working: if you divided cm³ by s, the rate is in cm³/s — not “cm³”, not “per second”, and not s/cm³. And before you substitute, check whether the units need converting: compare the units the question gives with the units it asks for — if the time is in minutes but the rate is wanted per second, change it to seconds first.
Rate graphs
A single mean rate hides the most interesting part of the story: the rate changes as the reaction runs. Plot the quantity of product formed (or reactant used) against time and the shape tells you everything:
- Steepest at the start — the rate is fastest at the beginning, when the concentrations of the reactants are highest.
- Gradient decreases — the reaction slows as the reactants are used up.
- Horizontal line — the reaction has finished: one of the reactants has run out, so no more product forms.
The same reaction seen two ways: product formed climbs to a plateau; reactant used up falls to one. The gradient is steepest at the start — the reaction is fastest when the reactant concentrations are highest — and zero when the reaction finishes.
Why the line goes flat — and how high
A rate curve flattens when the reaction finishes — and a reaction finishes the moment one reactant is completely used up. That reactant is the limiting reactant; whatever is left over is in excess. Crucially, the graph carries two separate pieces of information, and exam questions test them one at a time:
- How steep the curve is — the rate — is set by the conditions: concentration, temperature, surface area and any catalyst. Speed the reaction up and the curve rises more steeply (and reaches the plateau sooner).
- How high the plateau sits — the total amount of product — is set by the amount of the limiting reactant. Use more of it and the line levels off higher; use less and it levels off lower — no matter what you did to the rate.
So a faster reaction does not always make more product, and more product does not mean a faster reaction. Reading the steepness and the height as two independent things is one of the most discriminating graph skills in this topic — there’s a predict-the-curve trainer for it in section 2.
The limiting reactant is used up first, so it controls how much product can form; the reactant left over is in excess. It’s the same idea you met in C3 · Limiting Reactants — and it’s exactly why two runs with different rates can still finish at different heights on the graph.
When a question asks you to plot the data, marks sit in the mechanics before any chemistry happens:
- Choose scales so the plotted points fill at least half the grid in both directions — pick the scale first, before plotting anything. Use sensible steps (1, 2, 5 or 10 units per square), never awkward ones like 3 or 7.
- Label each axis with the quantity and its unit (“Volume of gas / cm³”, “Time / s”).
- Plot points accurately as small neat crosses — no “blobs” bigger than half a small square.
- Draw a single, smooth line (or curve) of best fit — not feathery sketching, not dot-to-dot, and never forced through an anomalous point.
The rate at a single moment — tangents
The mean rate over a whole experiment is one number, but the graph shows the rate constantly changing. To find the rate at one particular time, draw a tangent to the curve at that time — a straight line, ruled so it just touches the curve at that point and follows its direction — then measure the tangent’s gradient (slope). The steeper the tangent, the faster the reaction at that moment.
gradient = change in ychange in x = ΔyΔx
Everyone must be able to draw a tangent and use its slope to compare rates. Higher Higher Tier students must also calculate the gradient to give a numerical rate at that time — with the unit taken straight from the axes (cm³/s if the axes are cm³ and s).
Two habits separate full marks from dropped ones when measuring any gradient:
- Draw the largest triangle you reasonably can on the tangent — ideally spanning at least half the drawn line. Reading errors of half a square matter far less on a big triangle than on a tiny one, so a large triangle makes your gradient more accurate. Where the line crosses gridline intersections (or the axes) cleanly, those are ideal corner points — easy to read exactly.
- Annotate the coordinates you used. Mark the two points and write their values on the graph. Examiner reports note students who calculated a gradient but never showed which coordinates they read — making the working impossible to credit if the final answer slips.
📐 Tangent explorer — the rate at any moment
Magnesium + dilute hydrochloric acid, with the hydrogen collected in a gas syringe. Slide along the curve and watch the tangent, the triangle and the gradient calculation update. Notice when the reaction is fastest — and what the gradient becomes once it finishes.
Worked example — rate in mol/s Higher
Higher Tier questions sometimes want the rate in mol/s instead of cm³/s. You won’t be expected to know the conversion — the question gives you a factor — but you do need to draw (or read) a tangent first, then convert its gradient.
A tangent drawn to a graph of volume of oxygen against time has a gradient of 0.84 cm³/s. Under the conditions used, 1 cm³ of oxygen = 4.0 × 10⁻⁵ mol. Calculate the rate of reaction at that moment, in mol/s.
- The gradient of the tangent is the rate in cm³/s: 0.84 cm³/s.
- Convert each cm³ to moles with the factor given: rate = 0.84 × (4.0 × 10⁻⁵) mol/s.
- = 3.4 × 10⁻⁵ mol/s (2 s.f.). The unit follows the working: (cm³/s) × (mol/cm³) = mol/s.
- Mean rate — one number for a whole period: total quantitytotal time. Use it when the question says “calculate the mean rate between…” or gives a table of start/end values.
- Rate at an instant — the gradient of the tangent at that time. Use it when the question says “at 30 seconds” or “at this point”.
- Quoting a mean rate when the question asked for the rate at a time is a classic way to lose all the marks while doing correct arithmetic.
🧪 Exam-style questions
Calcium carbonate reacts with dilute hydrochloric acid. 36 cm³ of carbon dioxide is collected in the first 45 seconds. Calculate the mean rate of reaction over this period, in cm³/s.
Show answer
- mean rate = quantity of product formedtime taken = 3645. 1 mark
- = 0.8 cm³/s. 1 mark
- The unit comes straight from the working: cm³ divided by s gives cm³/s.
On a graph of volume of gas produced against time, when is the rate of reaction greatest?
A reaction loses mass as a gas escapes. Which set of units is correct for the rate of this reaction?
A tangent is drawn to a volume–time curve at t = 15 s. The tangent passes through the points (5, 18) and (35, 54), where x is in seconds and y in cm³. Calculate the rate of reaction at 15 s.
Show answer
- Δy = 54 − 18 = 36 cm³. 1 mark
- Δx = 35 − 5 = 30 s. 1 mark
- gradient = 3630 = 1.2 cm³/s. 1 mark
- Note how the two points span a wide run of the tangent — a big triangle — and both coordinates are stated in the working.
2Factors Affecting Rate & the Required Practical
Five factors can change the rate of a chemical reaction. You need to recall all five here, and explain them with collision theory in section 3:
| Factor | Increase it… |
|---|---|
| Concentration (of solutions) | …and the rate increases |
| Pressure (of reacting gases) | …and the rate increases |
| Surface area (of solid reactants) | …and the rate increases |
| Temperature | …and the rate increases |
| Catalyst present | Rate increases — and the catalyst is not used up (section 4) |
Each factor belongs to a particular state of matter: concentration is for solutions, pressure is for gases, and surface area is for solids. Writing “increase the concentration of the magnesium ribbon” tells the examiner the factors have been memorised without being understood — a solid doesn’t have a concentration.
Required Practical 5 — the effect of concentration
The required practical investigates how concentration affects rate, using three different methods of following a reaction. It is also the AQA specification’s named opportunity for developing a hypothesis — so before the method, write down what you expect and why: “increasing the concentration will increase the rate, because there are more particles in the same volume, so collisions are more frequent.”
Sodium thiosulfate solution reacts with dilute hydrochloric acid to make a precipitate of sulfur (a precipitate is an insoluble solid that forms when two solutions react), which slowly turns the mixture cloudy (the technical word is turbidity):
Na2S2O3(aq) + 2HCl(aq) → 2NaCl(aq) + S(s) + SO2(g) + H2O(l)
Read the state symbols: two clear solutions — thiosulfate (aq) and acid (aq) — react, and one of the products, the sulfur, comes out as a solid (s). It’s that fine solid sulfur, spread through the liquid, that clouds the mixture and hides the cross.
- Measure thiosulfate solution into a conical flask standing on a black cross drawn on paper.
- Add the acid, swirl, and start the stop clock.
- Look down through the solution; stop the clock the moment the cross is no longer visible.
- Repeat with different concentrations of thiosulfate, keeping everything else the same.
A shorter time means a faster rate — so rate is proportional to 1time.
React magnesium (or marble chips) with dilute hydrochloric acid and collect the gas in a gas syringe (or an upside-down measuring cylinder full of water). Record the volume every 10 seconds, repeat with a different acid concentration, and plot volume against time — exactly the curves from section 1. The steeper the initial gradient, the faster the rate.
The gas syringe is the neatest way to read a volume, but you can also collect the gas over water in an upturned measuring cylinder — the gas bubbles up and pushes the water down, and you read the volume off the cylinder’s scale:
Collecting a gas over water — an alternative to the gas syringe.
Two of the ways of following the rate in required practical 5. Method 1: time how long the black cross takes to vanish as the solution clouds. Method 2: read the gas volume off a syringe at regular times and plot the curve.
If a reaction gives off a gas, you can follow it on a balance. Stand the flask of reactants on the balance, plug the neck with cotton wool, and record the mass every few seconds. As gas escapes into the air the mass of the flask falls — and a faster fall means a faster reaction.
- The usual reaction is a carbonate with acid, e.g. CaCO3 + 2HCl → CaCl2 + H2O + CO2. The carbon dioxide leaves the flask, so the reading drops.
- The cotton wool plug earns marks of its own: it lets the gas escape (so the mass can change) while stopping any acid spray being thrown out as the mixture fizzes — spray leaving would make the mass fall too far.
- Mass only appears to be lost — no atoms are destroyed; the gas has simply left the flask. Seal the flask and the mass wouldn’t change at all. It’s the same open-vs-closed-system idea as C3 · Mass Changes Involving a Gas.
Following a reaction by loss of mass: a carbonate reacts with acid in a flask plugged with cotton wool. The CO₂ escapes into the air so the balance reading falls — while the cotton wool stops acid spray being lost with it.
Loss of mass works well for carbon dioxide (a fairly heavy gas), but badly for reactions that give off hydrogen (magnesium + acid). H2 is so light that even a lot of it barely moves the balance, so the change is lost in the noise — collect it in a gas syringe instead. Matching the method to the gas is itself a common exam point.
Variables — and the steps that make it valid
- Independent (you change it) — the concentration of sodium thiosulfate (or of the acid in method 2).
- Dependent (you measure it) — the time for the cross to disappear, or the volume of gas at set times.
- Control (keep the same) — the total volume (and so depth) of solution in the flask, the temperature (hold it steady with a water bath if the room is warm), the same flask and the same cross, and the same person judging when the cross has gone.
“Amount” is too vague to earn marks. Say exactly what you are changing, measuring or keeping the same — the mass, volume, concentration or number of moles. For example, write “the same volume of acid”, not “the same amount of acid”.
Examiner reports repeatedly show students naming the right experiment but missing the step that makes it work. For the disappearing cross, the decisive details are:
- Same depth of solution every time — you are looking down through the liquid, so a deeper mixture would hide the cross sooner and ruin the comparison. Keeping the total volume the same (topping up with water when diluting) keeps the depth the same.
- Measure the volumes accurately — use a measuring cylinder (not a beaker) to measure out the thiosulfate, acid and water. A measuring cylinder has a much finer scale — a better resolution — so each “same total volume” really is the same and the depth stays fair.
- State the observation, not an inference — timing stops when “the cross is no longer visible”. That is the observation. “The reaction finished” is an inference — and in fact it’s false: the reaction is still going; there’s just enough sulfur to block your view.
- Judging the end-point is the weakness — different people (or the same person on different runs) judge “no longer visible” differently. Improve it with the same observer every time, or remove the judgement entirely with a light sensor and data logger measuring the light transmitted through the flask.
- Safety — the reaction makes sulfur dioxide, a toxic gas: keep the room well ventilated (or use a fume cupboard at higher concentrations).
Try it below. Run all four concentrations and watch what happens to the time — then look at the pattern in 1/time.
🧪 Run required practical 5 — the disappearing cross
Choose a concentration of sodium thiosulfate, add the acid, and time how long the cross takes to vanish. The total volume is the same in every run — the diluted solutions are topped up with water — so the depth stays fair. Run all four and watch the pattern build.
| Conc / g/dm³ | Time / s | 1/t / s⁻¹ |
|---|
Time is not the rate — a shorter time means a faster reaction. 1/time behaves like the rate: watch it double when the concentration doubles.
1/t against concentration: a straight line through the origin — rate is directly proportional to concentration.
In the disappearing-cross experiment you measure a time, but the question usually asks about rate. They run opposite ways: the faster the reaction, the shorter the time. To turn times into something that behaves like a rate, calculate 1time — then a graph of 1/t against concentration gives a straight line through the origin, showing rate is directly proportional to concentration. (Doubling the concentration halves the time.)
Predict the curve — steepness vs plateau
This trainer drills the skill from section 1: for each change, decide separately what happens to the rate (how steep the curve is) and to the amount of product (how high the plateau sits). They don’t always move together.
📈 Predict the curve
A reaction is followed by the volume of gas it makes (the grey dashed curve). For each change, predict what happens to the steepness of the curve and the height of the plateau, then press Check.
Describing your measurements — the words examiners use
A handful of “working scientifically” words turn up again and again on rates papers, usually as a one-mark “which word describes…?” question with the others offered as distractors. They are easy marks once you can tell them apart.
- Repeatable — the same person, using the same method and equipment, gets similar results on repeating.
- Reproducible — different people, often with different equipment, get similar results. (The tougher test — it’s how science checks a claim.)
- Accurate — a measurement that is close to the true value.
- Precise — repeated measurements that are close together (a small spread). Not the same as accurate — tight readings can still all be off the true value.
- Valid — the experiment measures what it’s meant to and answers the question, because the other variables were controlled (a fair test).
- Resolution — the smallest change an instrument can detect — its finest scale division (a balance reading to 0.01 g has a higher resolution than one reading to 1 g).
- Anomalous result — a result that doesn’t fit the pattern. Examine it for a cause, and if it came from a poor measurement, leave it out of the mean.
🔗 Match the term to its meaning
Click a term, then click the definition that matches it. Correct pairs lock in green; a wrong pick flashes red.
Worked example — a mean that ignores the anomaly
A student repeats one concentration of the disappearing-cross run five times and records the times: 42 s, 44 s, 43 s, 58 s, 42 s. Calculate the mean time, to a sensible number of significant figures.
- Spot the anomaly first: four results cluster around 42–44 s, but 58 s is well out of line. It doesn’t fit the pattern, so it’s left out of the mean.
- Mean of the four that agree = 42 + 44 + 43 + 424 = 1714 = 42.75 s → 43 s (2 s.f.).
- Include the 58 s by mistake and the mean climbs to 45.8 s — the anomaly drags it up and makes the result less accurate. Always state which value you ignored, and why.
Spot the mistake in the apparatus
Examiners often draw a set-up with one thing wrong and ask why it won’t give good results. Find the fault in each diagram.
🔍 Spot the mistake
Each set-up has exactly one fault that would spoil the results. Tap the reason it won’t work.
Rates questions often ask you to identify the gas a reaction gives off. Three quick tests cover almost all of them:
- Carbon dioxide (carbonate + acid) — bubble it through limewater, which turns cloudy / milky.
- Hydrogen (metal + acid) — hold a lit splint to it for a squeaky pop.
- Oxygen (e.g. hydrogen peroxide decomposing) — it relights a glowing splint.
The full set of gas tests is C8 Chemical Analysis content — here you just need to recognise the gas your apparatus is collecting.
🧪 Exam-style questions
In the disappearing-cross investigation, which of these is the independent variable?
At one concentration, the cross took 80 seconds to disappear. Calculate 1/t for this run, in s⁻¹.
Show answer
- 1/t = 180 1 mark
- = 0.0125 s⁻¹ (= 1.25 × 10⁻² s⁻¹). 1 mark
- 1/t is used because it is proportional to the rate — a shorter time gives a bigger 1/t, matching a faster reaction.
Why must the total volume of solution be the same in every run of the disappearing-cross experiment?
Which change would remove the biggest source of error in the disappearing-cross method?
Three students, in different labs and using their own equipment, each measured the time for the cross to disappear at the same concentration and got very similar results. Which word best describes these results?
The rate of the reaction between marble chips and dilute hydrochloric acid was followed by measuring the volume of carbon dioxide with a gas syringe. Describe another method that could be used to follow the rate of this reaction.
Show a model answer
- Stand the flask on a balance (neck plugged with cotton wool) and record the mass at regular time intervals. 1 mark
- The mass falls as carbon dioxide escapes; a faster fall means a faster rate. 1 mark
- Allow: collect the gas over water in an inverted measuring cylinder and record the volume at set times. (The disappearing-cross method does not work here — this reaction doesn’t form a precipitate that clouds the solution.)
The diagram shows the gas syringe at one moment during the experiment. What volume of gas has been collected?
Plan an investigation, using sodium thiosulfate solution and dilute hydrochloric acid, to test the hypothesis that increasing the concentration of sodium thiosulfate increases the rate of reaction. Your plan should give valid results.
Show a model answer
A Level 3 plan needs a workable sequence that would actually produce the expected outcomes — not a heap of facts. Model answer:
- Measure 50 cm³ of sodium thiosulfate solution into a conical flask standing on a paper marked with a black cross.
- Add 10 cm³ of dilute hydrochloric acid, swirl once, and start the stop clock immediately.
- Look down through the solution and stop the clock when the cross is no longer visible; record the time.
- Repeat with at least four more concentrations of thiosulfate, diluting with water so the total volume is always the same (same depth).
- Keep the acid volume and concentration, the temperature, the flask, the cross and the observer the same — the control variables.
- Calculate 1/time for each run; if the hypothesis is right, 1/t will increase with concentration (a straight line through the origin shows direct proportionality). Repeat each concentration and use a mean, ignoring anomalies, to improve repeatability.
Structure tip — for any 6-mark plan: state the task (what you change and measure), give the sequence in a workable order, then say how the results answer the question. A plan that never says what to do with the times cannot reach the top level.
3Collision Theory & Activation Energy
Section 2 told you what speeds reactions up. Collision theory is the single idea that explains why — and it is worth learning properly, because almost every “explain” question in this topic is marked against it.
Chemical reactions can occur only when reacting particles collide with each other, and with sufficient energy. The minimum amount of energy that particles must have to react is called the activation energy.
So a reaction’s rate is set by two things: how often particles collide (the frequency of collisions), and what fraction of those collisions carry enough energy to count (at least the activation energy — the same activation energy you met on reaction profiles in C5). Speed either one up and the reaction gets faster.
Explaining each factor
| Factor increased | What changes for the particles | Effect |
|---|---|---|
| Concentration | More particles in the same volume | Collisions more frequent → rate increases |
| Pressure (gases) | Same particles squeezed into a smaller volume | Collisions more frequent → rate increases |
| Surface area (solids) | Smaller pieces → larger surface area : volume ratio, more particles exposed | Collisions more frequent → rate increases |
| Temperature | Particles move faster and carry more energy | Collisions more frequent and more energetic (a bigger fraction beat the activation energy) → rate increases a lot |
“More collisions” on its own is usually not enough for the mark. Left to run for long enough, even a slow reaction clocks up more collisions. What matters for rate is collisions per second — so write “collisions are more frequent” (or “more collisions per second”). The same discipline applies to energy: it’s not that particles “hit harder”, it’s that a greater proportion of collisions have energy ≥ the activation energy — these are often called successful collisions. Put the two together and you have the phrase that reliably earns the mark: rate depends on the frequency of successful collisions — how many collisions with energy ≥ the activation energy happen per second. A bare “number of successful collisions” still names no time period, so on its own it can’t describe a rate.
Concentration, pressure and surface area work through frequency only. Temperature is the only factor that works through both frequency and energy — and the energy effect dominates, which is why a small temperature rise can have a dramatic effect on rate. An “explain the effect of temperature” answer that only mentions faster-moving particles colliding more often is doing half the job: the second mark almost always needs “more particles collide with energy greater than or equal to the activation energy, so more collisions are successful”.
Surface area — smaller pieces, bigger ratio
Cutting a solid into smaller pieces doesn’t change how much of it there is — it changes how much of it is on the outside. You need to describe this in terms of the surface area to volume ratio: smaller pieces have a larger surface area to volume ratio, so more particles are exposed for collisions and the rate increases. The extreme case is a powder, which reacts far faster than the same mass in lumps.
Cut a 2 cm cube into eight 1 cm cubes and the volume is unchanged — but every cut exposes two new faces, doubling the surface area and the SA:V ratio. More exposed particles → more frequent collisions → faster reaction.
Proportionality — the simple maths of collisions
For factors that work through frequency alone, the maths is direct: double the concentration and there are twice as many particles in the same volume, so collisions are twice as frequent and the rate roughly doubles — exactly the straight-line-through-the-origin pattern from the required practical in section 2. Temperature does not follow this simple proportionality: because it also boosts the fraction of successful collisions, a mere 10 °C rise can roughly double the rate of many reactions.
rate ∝ concentration and rate ∝ 1time
The symbol ∝ means “is directly proportional to” — double one side and the other doubles too. It’s the compact way to write the straight-line-through-the-origin patterns from the required practical in section 2.
Watch all of this happen below. Choose a scenario — a solid in a solution, or two gases — then flip the factors that apply to it and watch how each one changes the rate of successful collisions.
🔬 Collision theory — explore the five factors
Pick a scenario, then explore the factors. Particles move and collide; a collision with at least the activation energy reacts — a ✦ spark and a purple product. Each factor only appears for the state of matter it actually applies to, so the science stays true. Watch the rate of successful collisions respond.
All five factors are at their baseline. Increase any one and watch the rate of successful collisions respond — and notice why it changes.
🧪 Exam-style questions
Which factor increases the rate of reaction by increasing both the frequency and the energy of collisions?
Powdered calcium carbonate reacts faster with acid than the same mass of marble chips because the powder has…
The concentration of an acid is doubled (everything else unchanged), and the initial rate of its reaction with magnesium roughly doubles. Why?
Increasing the pressure of a reacting gas increases the rate of reaction. Which statement explains why?
Explain, using collision theory, why increasing the temperature increases the rate of a reaction.
Show a model answer
- At a higher temperature the particles move faster… 1 mark
- …so collisions are more frequent (more collisions per second). 1 mark
- The particles also have more energy, so a greater proportion of collisions have energy ≥ the activation energy… 1 mark
- …so there are more successful collisions per second, and the rate increases. 1 mark
Examiner note — this is the full causal chain: faster particles → more frequent collisions; more energy → bigger fraction over Eₐ. Answers that say only “more collisions”, or that describe (“the reaction speeds up”) without explaining, stall at half marks.
4Catalysts
The fifth way to speed up a reaction is the cleverest: add a substance that makes the reaction faster without being used up itself.
A catalyst changes the rate of a chemical reaction but is not used up during the reaction. Because it isn’t a reactant, it is not included in the chemical equation. Different reactions need different catalysts — for example, enzymes act as catalysts in biological systems.
How do they do it? A catalyst provides a different pathway for the reaction — one with a lower activation energy. The particles themselves are no faster and no more energetic than before; the bar has simply been lowered, so a greater proportion of the collisions that were already happening now have enough energy to react. More successful collisions per second — a faster rate. You can picture it on a reaction profile — the same diagram first introduced with the reaction profiles in C5 — where the catalysed pathway climbs a noticeably lower hump:
- It does not give the particles more energy — it lowers the activation energy by providing a different pathway. (Raising the temperature does the reverse: it leaves the barrier where it is but gives the particles more energy. A catalyst lowers the bar; heating lifts the particles over it.)
- It does not change the overall energy change of the reaction, and it does not make more product — only faster product.
- It is not used up — but avoid saying it “doesn’t take part”. It does take part (that’s how it provides the new pathway); it is regenerated by the end, which is why it never appears in the equation.
Spotting a catalyst in exam data
Questions rarely say “X is a catalyst” — they make you identify it from its fingerprints. Look for a substance that:
- speeds the reaction up (gas produced faster, time shorter, initial gradient steeper);
- has the same mass at the end as at the start (not used up) — often it can be filtered out and reused;
- does not appear in the chemical equation for the reaction.
A classic demonstration is the decomposition of hydrogen peroxide, which is very slow on its own but fizzes vigorously with a little manganese(IV) oxide powder:
2H2O2(aq) → 2H2O(l) + O2(g)
(MnO2 catalyst — written above the arrow, never in the equation.)
You don’t need to memorise lists of catalysts, but one pattern helps you identify them in data questions: transition metals and their compounds are very often the catalyst (manganese(IV) oxide above; iron, nickel, platinum, copper and their ions elsewhere) — acting as catalysts is one of the characteristic properties of the transition elements you met in C1. So when a question shows several substances and asks which is most likely to be the catalyst, a transition-metal compound or ion is the safe bet over a main-group salt. (A classic class investigation compares different metal salts added to hydrogen peroxide.)
Catalysts matter to industry for a very practical reason: a faster reaction at a lower temperature means less energy bought from burning fuel — lower costs and less environmental impact. Enzymes do the same job in living things (and in industry too — they are the catalysts in fermentation).
🧪 Exam-style questions
A catalyst increases the rate of a reaction by…
0.5 g of manganese(IV) oxide is added to hydrogen peroxide solution. The mixture fizzes rapidly. When the reaction has finished, the manganese(IV) oxide is filtered off, dried and reweighed. What mass is found?
Which of these does a catalyst change?
A student adds a small amount of each of these substances to separate samples of hydrogen peroxide solution. Which is most likely to act as a catalyst for its decomposition?
On a reaction profile, what does adding a catalyst change, and what stays the same?
Show a model answer
- Changes: the height of the “hump” — the catalysed pathway has a lower activation energy. 1 mark
- Stays the same: the energy levels of the reactants and products — so the overall energy change is identical. 1 mark
5Reversible Reactions
So far every reaction has gone one way: reactants → products, done. But in some reactions the products can react together to re-form the original reactants. These are reversible reactions, and they get their own arrow:
A + B ⇌ C + D
The double half-arrow ⇌ means the reaction goes in both directions at the same time: A and B react to make C and D (the forward reaction), while C and D react to re-form A and B (the reverse reaction). Which direction is favoured can be controlled by changing the conditions — temperature, pressure or concentration.
A classic example is heating ammonium chloride. Warm it and the white solid breaks down into two gases; let the gases cool and they recombine into the white solid — which is why a heated tube of ammonium chloride grows a white ring further up, where the gases meet cooler glass:
ammonium chloride ⇌ ammonia + hydrogen chloride
NH4Cl(s) ⇌ NH3(g) + HCl(g)
White ammonium chloride solid sits in the bottom of the tube.
Heating pushes the reaction forward (decomposition) at the hot bottom of the tube; higher up, the cooler glass pushes it back (recombination) — the same reversible reaction running both ways at once.
Energy changes in reversible reactions
If a reversible reaction is exothermic in one direction, it is endothermic in the opposite direction — and the same amount of energy is transferred in each case. Energy is conserved: whatever the forward reaction gives out, the reverse reaction must take back in.
Some questions give the energy change as ΔH with a sign instead of the words “exothermic” or “endothermic”. A negative ΔH (e.g. ΔH = −45 kJ/mol) means the forward reaction is exothermic; a positive ΔH means it is endothermic. The reverse reaction is then the opposite, transferring the same amount of energy. (ΔH notation is strictly A-level — at GCSE you just read the sign; there’s more with the reaction profiles in C5.)
You can see this in the ammonium chloride reaction above: the forward direction (decomposition) only happens while you keep heating it, so it is endothermic — which means the reverse change (the gases recombining into the white ring on the cooler glass) must be exothermic, releasing exactly the same amount of energy.
🧪 Exam-style questions
What does the symbol ⇌ in an equation tell you?
The forward direction of a reversible reaction is exothermic and transfers 58 kJ of energy. How much energy is transferred by the reverse reaction?
Show answer
- 58 kJ — taken in (endothermic). 1 mark
- If a reversible reaction is exothermic one way, it is endothermic the other way, and the same amount of energy is transferred in each case.
For a reversible reaction, the forward direction has ΔH = −45 kJ/mol. Which statement is correct?
Heating ammonium chloride in a test tube produces a white ring of ammonium chloride higher up the tube. Explain why.
Show a model answer
- Heating decomposes the ammonium chloride into ammonia and hydrogen chloride gases (the forward, endothermic direction). 1 mark
- The gases rise to a cooler part of the tube, where the reverse reaction happens: they recombine into solid ammonium chloride — the white ring. 1 mark
- One experiment, both directions: changing the conditions (here, temperature) changes which way a reversible reaction is favoured.
6Dynamic Equilibrium
Run a reversible reaction in an open beaker and the gaseous products drift away — the reverse reaction never gets its ingredients. But seal the same reaction into a closed system (“apparatus which prevents the escape of reactants and products”) and something remarkable happens.
At first the forward reaction is fast — the flask is full of reactants — and the reverse reaction can barely happen, because there are hardly any products to react. As reactants are used up, the forward rate falls; as products accumulate, the reverse rate rises. Eventually the two rates meet:
When a reversible reaction occurs in apparatus which prevents the escape of reactants and products, equilibrium is reached when the forward and reverse reactions occur at exactly the same rate.
Approaching equilibrium in a closed system: the forward rate falls as reactants are used up, the reverse rate climbs as products build, and at equilibrium the two are exactly equal — both reactions still running, cancelling each other out.
The same approach to equilibrium looks just as clear if we plot the concentrations against time instead of the rates:
Now plotted as concentration against time. The reactants fall and the products rise until, at equilibrium, every concentration goes constant — the lines flatten together. But notice they settle at different heights: at equilibrium the amounts are constant, not equal. How high each one settles depends on where the position of equilibrium lies.
That is why it is called dynamic equilibrium: nothing has stopped. Both reactions are still running, but because they run at the same rate, the concentrations of all the substances stay constant. From the outside the mixture looks frozen; at the particle level it is anything but.
- “At equilibrium the reaction stops.” No — both reactions continue. The rates are equal, so there is no further overall change. Concentrations are constant, not static because nothing is happening.
- “At equilibrium there are equal amounts of reactants and products.” No — the amounts are constant, not equal. An equilibrium can sit far to the right (mostly products) or far to the left (mostly reactants). Where it sits is called the position of equilibrium — and moving that position is the whole of section 7.
Equilibrium can only be reached in a closed system. If a product escapes — a gas bubbling out of an open flask — the reverse reaction is starved of its reactant, the reverse rate can never catch up with the forward rate, and the reaction simply runs to completion. (The heated ammonium chloride tube in section 5 only forms its white ring because the tube traps the gases long enough for them to recombine.)
Some equilibria involve a coloured substance, and the colour reveals where the position lies. A classic is hydrogen + iodine ⇌ hydrogen iodide, H2(g) + I2(g) ⇌ 2HI(g): iodine vapour is purple, while H2 and HI are colourless. As the forward reaction proceeds the mixture fades to a paler purple — but it never goes colourless, because at equilibrium some I2 always remains (the reaction doesn’t go to completion). “Paler, not colourless” is the mark-scheme distinction.
The colour reveals the position
Drag the slider to move the position of equilibrium left or right, and watch the colour change as the amounts change.
H2(g) + I2(g) ⇌ 2HI(g)
I2 vapour is purple; H2 and HI are colourless — so the colour you see tracks the amount of I2. However far right the position lies, a little I2 always remains: paler, never colourless.
🧪 Exam-style questions
Which statement is true for a reversible reaction at equilibrium?
Why can equilibrium only be reached in a closed system?
At equilibrium, the concentrations of all reactants and products are…
Colourless hydrogen and purple iodine vapour are sealed in a tube and left to reach equilibrium: H2(g) + I2(g) ⇌ 2HI(g). What is the colour of the mixture at equilibrium?
7Changing Equilibrium Conditions Higher
The relative amounts of reactants and products at equilibrium — the position of equilibrium — depend on the conditions. Change the conditions and the position moves. Higher Tier students must predict which way, using one beautifully simple rule:
If a system is at equilibrium and a change is made to any of the conditions, the system responds to counteract the change.
Whatever you do to an equilibrium, it shifts in the direction that partly undoes what you did. Add more of something and the equilibrium uses it up; heat it and the equilibrium absorbs the heat; increase the pressure and the equilibrium reduces the number of gas molecules. You are only asked for qualitative predictions, and the question will always give you the information you need — the equation, and which direction is exothermic.
The three changes
1 · Concentration. If the concentration of one substance is changed, the system is no longer at equilibrium, and the concentrations of all the substances change until equilibrium is reached again:
- Increase a reactant’s concentration → the equilibrium shifts to use it up → more products form until equilibrium is re-established.
- Decrease a product’s concentration (e.g. keep removing it) → the equilibrium shifts to replace it → more reactants react. This is how industry drags an equilibrium toward the product it wants.
2 · Temperature. Heat is “counteracted” by the direction that absorbs it — the endothermic direction; cooling favours the exothermic direction:
| Change | Favoured direction | Result |
|---|---|---|
| Increase temperature | Endothermic direction (absorbs the extra heat) | If the forward reaction is endothermic → more products. If it’s exothermic → fewer products. |
| Decrease temperature | Exothermic direction (releases heat to replace what was lost) | If the forward reaction is exothermic → more products. If it’s endothermic → fewer products. |
3 · Pressure (gaseous equilibria only). Before the rule, it is worth remembering what pressure actually is.
Pressure = force ÷ area. For a gas, the force is its molecules constantly striking the walls of the container, and the area is the inside surface of that container — set by its volume. Pack more gas molecules into the space (or squeeze the same molecules into a smaller volume) and the walls are struck more often, so the pressure rises.
So the number of gas molecules is what sets the force on the walls — and that is the one lever the equilibrium can pull. It cannot change the size of the container you have chosen, but it can shift to the side with fewer gas molecules: fewer molecules → fewer collisions with the walls → lower pressure.
So the system counteracts an increase in pressure by shifting toward the side with fewer gas molecules:
- Increase the pressure → equilibrium shifts toward the side with the smaller number of gas molecules, as shown by the symbol equation.
- Decrease the pressure → equilibrium shifts toward the side with the larger number of gas molecules.
- Count the molecules from the balanced equation’s coefficients — e.g. in N2 + 3H2 ⇌ 2NH3 there are 4 molecules on the left and 2 on the right.
Le Chatelier questions are marked on the complete chain of reasoning, not the conclusion alone. Build every answer in three links:
- State what the change is in the system’s terms (“temperature increased”, “4 gas molecules on the left, 2 on the right”).
- Apply the principle: the equilibrium shifts in the direction that counteracts it — naming that direction and why (“…shifts in the endothermic direction, which here is the reverse reaction, to absorb the extra heat”).
- Finish with the effect the question actually asked about: “…so the amount/yield of NH₃ at equilibrium decreases”. Stopping one step short — naming the shift but never the consequence — is one of the most common ways to lose the final mark.
Now try it. The simulator below builds the ammonia equilibrium from scratch — press Start to watch the forward and reverse reactions reach a dynamic balance, then change one condition at a time and read the examiner-style reasoning for every shift.
⚖️ The ammonia equilibrium, live
Remember: the forward reaction is exothermic. Press Start to reach equilibrium, then change one condition at a time and read what happens.
Only reactants so far — 6 N₂ and 18 H₂ bouncing around. Press Start to let them react and watch the mixture settle into equilibrium.
- Mixing up rate and position. A catalyst speeds up both directions equally: equilibrium is reached sooner, but the position (and yield) is unchanged. Don’t claim a catalyst increases yield.
- Counting molecules wrong for pressure. Count gas molecules only, using the balanced equation’s coefficients. If both sides have the same number of gas molecules, changing the pressure does not shift the position at all.
- Forgetting which direction is exothermic. The question always tells you (or gives data to work it out) — quote it in your answer: “the forward reaction is exothermic, so the endothermic direction is the reverse…”
Reading an equilibrium-yield graph
Higher Tier questions often give a graph of the % yield of a product against pressure, with a separate curve for each temperature. Read it with the same Le Chatelier logic, in two directions:
- Along a curve (pressure rising) — for N2 + 3H2 ⇌ 2NH3 the yield rises with pressure, because higher pressure favours the side with fewer gas molecules (2 on the right, not 4 on the left).
- Between curves (temperature) — the lower the temperature, the higher the yield, because the forward reaction is exothermic and cooling favours it.
- So the best yield sits in the top-right corner: high pressure, low temperature. Why industry doesn’t simply operate there is the very next point.
Yield of ammonia against pressure at three temperatures.
This is where the two halves of C6 finally meet. Take a forward reaction that is exothermic and makes fewer gas molecules, like N2 + 3H2 ⇌ 2NH3. The highest equilibrium yield wants a low temperature (favouring the exothermic forward direction) and a high pressure (favouring the 2-molecule side). But a low temperature also means a slow rate — you could wait days for that yield.
So real processes settle on a compromise: a moderate temperature (some yield given up to keep the rate usable), a catalyst (which speeds the reaction up without moving the position, so it lifts the rate at no cost to the yield), and a high pressure where the equation — and the expense of the equipment — allow it. And because each pass through the reactor only converts some of the mixture, the unreacted reactants are separated off and recycled back through — so very little is wasted, and the overall yield of the process is high even when a single pass is modest. If a question asks why a chosen temperature isn’t the one that gives the best yield, the answer is always this rate–yield trade-off.
The industrial optimisation and compromise, and reading an equilibrium-yield graph is a Higher-Tier data skill for everyone. The full industrial detail — recycling the unreacted gases, and choosing the conditions for a named process such as the Haber process — belongs to the Using Resources topic (4.10), and is Triple Chemistry only. It’s shown here, kept generic, just to bring the rate–yield idea to life — Combined Science students don’t need the named-process specifics.
🧪 Exam-style questions Higher
In the equilibrium ICl(l) + Cl2(g) ⇌ ICl3(s), the forward reaction is exothermic. What happens to the amount of ICl3 if the temperature is increased?
For the equilibrium 2SO2(g) + O2(g) ⇌ 2SO3(g), what is the effect of increasing the pressure?
Ethanol is made industrially by the equilibrium C2H4(g) + H2O(g) ⇌ C2H5OH(g). The ethanol is condensed and removed as it forms. Why does this increase the yield?
Hydrogen for industry is made using the equilibrium CH4(g) + H2O(g) ⇌ CO(g) + 3H2(g). The forward reaction is endothermic. Predict and explain the effect on the equilibrium yield of hydrogen of (a) increasing the temperature, and (b) increasing the pressure.
Show a model answer
(a) Increasing the temperature:
- The equilibrium shifts in the endothermic direction to absorb the extra heat — here that is the forward direction. 1 mark
- So the equilibrium shifts right and the yield of hydrogen increases. 1 mark
(b) Increasing the pressure:
- There are 2 gas molecules on the left and 4 on the right; higher pressure favours the side with fewer gas molecules — the left. 1 mark
- So the equilibrium shifts left and the yield of hydrogen decreases. 1 mark
Examiner note — both parts use the same three links: the change, the counteracting direction with its reason, and the effect on yield. Notice the question gave the thermal information (“the forward reaction is endothermic”) — quote it, don’t guess it.
Ammonia is made industrially from the equilibrium N2(g) + 3H2(g) ⇌ 2NH3(g). The forward reaction is exothermic. Explain how the temperature, the pressure and a catalyst should be chosen so that ammonia is made economically.
Show a model answer
This is the most common C6 six-marker. A full answer weighs yield against rate and cost for each condition — not just “use a high pressure”.
Temperature
- A low temperature gives a higher yield, because it favours the exothermic forward reaction… 1 mark
- …but a low temperature also makes the rate slow, so a moderate (compromise) temperature is used — some yield is given up to keep the rate usable. 1 mark
Pressure
- A high pressure gives a higher yield, because it favours the side with fewer gas molecules (2 on the right, not 4), and it also speeds the reaction up… 1 mark
- …but very high pressures need expensive, strong equipment and are hazardous, so a high but not extreme pressure is the compromise. 1 mark
Catalyst and recycling
- A catalyst speeds the reaction up without moving the position of equilibrium — it raises the rate at no cost to the yield (and lets a lower temperature be used). 1 mark
- The unreacted nitrogen and hydrogen are recycled, so little is wasted and the overall process is economical. 1 mark
Examiner note — a Level-3 answer pairs each condition with both its effect on the yield and the rate-or-cost reason it has to be a compromise. Listing the conditions without the trade-off caps the marks.
- Calculating rates: mean rate = quantity of reactant used (or product formed)time taken; units g/s or cm³/s (mol/s H). The graph is steepest at the start; draw a tangent for the rate at an instant, and H calculate its gradient.
- Factors & RP5: rate increases with higher concentration, gas pressure, surface area, temperature and a catalyst. Follow a reaction by gas volume (syringe or over water), loss of mass on a balance, or a colour/turbidity change (the disappearing cross). Required practical 5 links concentration to rate — time isn’t rate, so use 1/time. The graph’s steepness is the rate; its plateau height is set by the amount of the limiting reactant.
- Collision theory: particles react only when they collide with at least the activation energy. Factors raise the frequency of collisions; temperature also makes a bigger fraction successful; smaller pieces react faster because of a larger surface area : volume ratio.
- Catalysts: speed up a reaction by providing a different pathway with a lower activation energy, without being used up — so they are not in the equation and change neither the yield nor the overall energy change. Enzymes are biological catalysts.
- Reversible reactions (⇌): the products can react to re-form the reactants; if the reaction is exothermic one way it is endothermic the other, transferring the same amount of energy each way (e.g. heating ammonium chloride).
- Dynamic equilibrium: reached in a closed system when the forward and reverse reactions occur at the same rate (not zero). Concentrations stay constant — constant does not mean equal.
- Changing conditions H: by Le Chatelier’s principle the position shifts to counteract a change — raise temperature → endothermic direction; raise pressure → side with fewer gas molecules; add a reactant or remove a product → towards products. Real processes balance the rate–yield compromise.