Collision theory: why some reactions are fast and others slow
By the end of this lesson you will be able to
- Define rate of reaction and calculate a mean rate with correct units (cm³ s⁻¹, g s⁻¹, mol s⁻¹).
- State the three conditions of collision theory and define activation energy.
- Explain, at the particle level, how concentration, gas pressure, surface area, temperature and the nature of the reactants change the rate.
- Calculate how surface area changes when a solid is broken up, and estimate the effect of a temperature change with the “doubling per 10 °C” rule of thumb.
- Distinguish between the two effects of temperature — more frequent collisions and more energetic collisions — and evaluate which matters more.
You met this before
- Particles in liquids and gases are in constant random motion; raising the temperature raises their average kinetic energy, so they move faster (Year 1 particle model; Unit 3 kinetic theory).
- In a chemical reaction bonds in the reactants break and new bonds form in the products; atoms are rearranged, never created or destroyed (Year 3 Energy and chemical change).
- Breaking a bond takes in energy (endothermic); making a bond gives out energy (exothermic). ΔH is negative for an exothermic reaction (Year 3; Unit 8 energetics).
- Concentration is the amount of solute in a volume of solution: c = n ÷ V, in mol dm⁻³ (Unit 2).
- A carbonate reacts with an acid to give a salt, water and carbon dioxide: CaCO₃(s) + 2HCl(aq) → CaCl₂(aq) + H₂O(l) + CO₂(g) (Year 3 Kinds of reaction).
- One mole of any gas occupies about 24 dm³ (24 000 cm³) at room temperature and pressure (Unit 2).
A solution contains 0.050 mol of hydrochloric acid in 250 cm³. Calculate its concentration in mol dm⁻³, and state what happens to the average speed of the water molecules and ions if the solution is warmed from 20 °C to 40 °C.
Reveal
V = 250 cm³ ÷ 1 000 = 0.250 dm³
c = n ÷ V
c = 0.050 mol ÷ 0.250 dm³
c = 0.20 mol dm⁻³
Warming raises the average kinetic energy of the particles, so their average speed increases (from 293 K to 313 K is a 7 % rise in kelvin temperature). This lesson is about what those faster particles do when they meet.
Revise it in full: Year 3 · Energy and chemical change · Lesson 3 — Kinds of reaction · Year 3 · Energy and chemical change · Lesson 4 — Energy in and out: exothermic and endothermic change · Year 3 · Solutions, acids and bases · Lesson 2 — How much is dissolved: concentration
The puzzle
A spoonful of flour on a kitchen shelf will sit there for a year without changing. The same flour, blown into a cloud inside a closed mill and met by a single spark, can burn so fast that the building explodes. Grain silos, sugar refineries and coal mines have all been destroyed this way.
Nothing about the flour changed — same substance, same oxygen, same room temperature. What changed was how the particles meet. A diamond is pure carbon and so is charcoal; both burn in oxygen, yet one glows in a braai for hours and the other lasts for billions of years. Why does the speed of a reaction depend on so much more than what is reacting?
Predict which will produce carbon dioxide faster in the same volume of the same acid: one 2 g marble chip, or 2 g of marble powder. Then predict what happens to the total volume of carbon dioxide made when all the marble has reacted in each case.
Compare your answer
The powder reacts much faster, because far more of its particles are on a surface where acid particles can hit them. But the total volume is the same: 2 g of calcium carbonate is 0.020 mol and makes 0.020 mol of carbon dioxide (about 480 cm³) whichever form it is in.
Rate is how fast; the amount of product depends on how much reactant there is. Keeping those two ideas apart is the first habit of this unit.
The science — structured notes
1.1 What “rate of reaction” means, and its units
The rate of a reaction is the change in the amount (or concentration) of a reactant or product per unit time. It tells you how fast reactants are used up or products are made, not how much is made in total.
mean rate of reaction = change in amount of product (or reactant)time taken
The unit is whatever you measured, divided by a unit of time. A gas syringe gives cm³ s⁻¹; a balance gives g s⁻¹; a calculation in moles gives mol s⁻¹; a change of concentration gives mol dm⁻³ s⁻¹. Rate is almost never measured directly — it is calculated from a measured change and a measured time, which is why examiners do not accept “rate” as a dependent variable on its own.
Marble chips react with hydrochloric acid. The gas syringe collects 36 cm³ of carbon dioxide in the first 30 s. Calculate the mean rate over the first 30 s in cm³ s⁻¹ and in mol s⁻¹ (1 mol of gas occupies 24 000 cm³ at room conditions).
Rate in cm³ s⁻¹. mean rate = volume ÷ time
mean rate = 36 cm³ ÷ 30 s
mean rate = 1.2 cm³ s⁻¹
Moles of gas. n = V ÷ 24 000 cm³ mol⁻¹
n = 36 cm³ ÷ 24 000 cm³ mol⁻¹
n = 1.5 × 10⁻³ mol
Rate in mol s⁻¹. 1.5 × 10⁻³ mol ÷ 30 s
mean rate = 5.0 × 10⁻⁵ mol s⁻¹
A flask of acid and marble chips on a balance loses 0.66 g in the first 2.0 minutes as carbon dioxide escapes. Calculate the mean rate of mass loss in g s⁻¹, and the mean rate of carbon dioxide production in mol s⁻¹ (M(CO₂) = 44.0 g mol⁻¹).
Reveal answer
Time = 2.0 × 60 = 120 s
Rate of mass loss = 0.66 g ÷ 120 s
Rate of mass loss = 5.5 × 10⁻³ g s⁻¹
n(CO₂) = 0.66 g ÷ 44.0 g mol⁻¹
n(CO₂) = 0.015 mol
Rate = 0.015 mol ÷ 120 s
Rate = 1.25 × 10⁻⁴ mol s⁻¹ (1.3 × 10⁻⁴ to two significant figures, matching the data)
1.2 Collision theory: three conditions for a reaction
Particles can only react if they meet. Collision theory says that a reaction happens when reactant particles collide, and that only some collisions are successful. A collision leads to reaction only if:
- the particles collide — they must actually meet (in a solid, only the surface particles can be hit);
- with enough energy — the combined kinetic energy along the line of impact must be at least the activation energy, Ea, the minimum energy needed to start breaking the bonds that must break;
- with a suitable orientation — the parts of the particles that react must face each other.
Most collisions fail. In a gas at room temperature each molecule collides billions of times a second, yet most reactions take seconds or hours, because only a tiny fraction of those collisions have enough energy. So the rate depends on two things: how often particles collide (collision frequency) and what fraction of those collisions are successful.
Hydrogen and oxygen can be mixed in a balloon at room temperature and nothing happens for years; a spark makes the mixture explode. Use the idea of activation energy to explain both observations.
Reveal answer
At room temperature the molecules collide constantly, but almost none of the collisions have energy at least equal to the activation energy needed to begin breaking the H–H and O=O bonds, so there are almost no successful collisions and the rate is negligible.
The spark gives a few molecules enough energy to react. The reaction is strongly exothermic, so the energy released heats nearby molecules above the activation energy, they react and release more energy — a chain that spreads through the whole mixture in a fraction of a second.
1.3 Concentration and gas pressure: more particles in the same space
Increasing the concentration of a solution puts more reactant particles into each cubic decimetre. The particles move at the same average speed (same temperature), so they collide more often: more collisions per second, and therefore more successful collisions per second, even though the fraction that succeed is unchanged.
For a gas, pressure plays the same role. At constant temperature, doubling the pressure doubles the number of molecules in each cubic decimetre (Unit 3: p = (n ÷ V) RT), so the collision frequency and the rate rise. That is one reason industrial gas reactions — the Haber process of Lesson 5 — run at high pressure.
“Doubling the concentration of the acid doubles the rate because each collision now has twice as much energy.”
Reveal answer
False. At the same temperature the particles have the same average kinetic energy, so the fraction of collisions with at least Ea is unchanged. What doubles is the number of particles per unit volume, and so the frequency of collisions — and with it the number of successful collisions per second.
1.4 Surface area: only the outside of a solid can be hit
When one reactant is a solid, the other particles can only collide with the particles on its surface. Breaking the solid into smaller pieces exposes particles that were hidden inside, so the surface area rises, collisions become more frequent and the rate rises. The mass, and so the total amount of product, does not change.
The effect is large because surface area grows as the pieces get smaller. For a cube of side L, the surface area is 6L²; cutting it into smaller cubes keeps the volume but multiplies the surface.
A cube of marble has sides of 2.0 cm. Calculate its surface area, and the total surface area after it is cut into cubes of side 1.0 cm and then of side 0.50 cm.
One cube. SA = 6 × L²
SA = 6 × (2.0 cm)²
SA = 24 cm²
Side 1.0 cm. Number of cubes = (2.0 ÷ 1.0)³ = 8
SA = 8 × 6 × (1.0 cm)²
SA = 48 cm²
Side 0.50 cm. Number of cubes = (2.0 ÷ 0.50)³ = 64
SA = 64 × 6 × (0.50 cm)²
SA = 96 cm²
Pattern. Halving the side doubles the total surface area: SA is inversely proportional to the side length for a fixed volume.
A sugar cube of side 1.0 cm is ground into tiny cubes of side 0.010 cm (0.1 mm), the size of dust in a refinery. Calculate how many times larger the surface area becomes, and use collision theory to explain why sugar dust in air can explode while a sugar cube cannot.
Before. SA = 6 × (1.0 cm)²
SA = 6.0 cm²
Number of dust cubes. (1.0 ÷ 0.010)³ = 1.0 × 10⁶
After. SA = 1.0 × 10⁶ × 6 × (0.010 cm)²
SA = 1.0 × 10⁶ × 6.0 × 10⁻⁴ cm²
SA = 600 cm²
Ratio. 600 ÷ 6.0 = 100 times larger
Meaning. Every dust particle is surrounded by air, so oxygen molecules collide with a hundred times more sugar surface per second; the heat from one burning grain reaches its neighbours at once and the whole cloud burns in a fraction of a second. In February 2008 a sugar-dust explosion destroyed part of a refinery in Georgia, USA, and killed fourteen workers.
A 1.0 cm cube of zinc is cut into cubes of side 0.25 cm. Calculate the new total surface area and state by what factor the initial rate of reaction with excess acid would be expected to increase, if rate is proportional to surface area.
Reveal answer
Number of cubes = (1.0 ÷ 0.25)³ = 64
SA = 64 × 6 × (0.25 cm)²
SA = 24 cm² (from 6.0 cm² before)
Factor = 24 ÷ 6.0 = 4
The initial rate would rise about four-fold; the total volume of hydrogen at the end is unchanged because the mass of zinc is the same.
1.5 Temperature: faster particles, and far more of them above Eₐ
Raising the temperature does two things, and examiners want both:
- The particles move faster, so they collide more often. This is a small effect: from 20 °C to 30 °C the average speed rises by less than 2 %.
- A much greater proportion of collisions have energy at least equal to Ea. This is the big effect. Because only the fastest particles have enough energy, a small rise in the average pushes many more of them over the threshold (Lesson 3 draws the energy distribution that shows why).
For many reactions near room temperature, a 10 °C rise roughly doubles the rate. This is a rule of thumb, not a law: the factor depends on the activation energy and is often between 1.5 and 3.
At 20 °C a reaction takes 160 s to produce 50 cm³ of gas. Assuming the rate doubles for every 10 °C rise, estimate the time at 50 °C, and the mean rate at each temperature.
Number of 10 °C steps. (50 − 20) ÷ 10 = 3
Rate factor. 2³ = 8
Time. 160 s ÷ 8 = 20 s
Mean rate at 20 °C. 50 cm³ ÷ 160 s = 0.31 cm³ s⁻¹
Mean rate at 50 °C. 50 cm³ ÷ 20 s = 2.5 cm³ s⁻¹
Time and rate are inversely related: eight times the rate, one eighth of the time.
Milk left out at 30 °C goes sour in about half a day. Using the doubling rule, estimate how long it would take in a fridge at 5 °C, and explain the principle of refrigeration in terms of collisions.
Reveal answer
Steps of 10 °C from 30 °C down to 5 °C = 2.5
Rate factor = 22.5 ≈ 5.7 (slower)
Time ≈ 0.5 day × 5.7 ≈ 3 days
(Real milk usually keeps for longer than this in a fridge: the doubling rule is only a rough guide, and bacterial growth slows especially steeply near 5 °C. The estimate still shows the right idea.) Cooling lowers the kinetic energy of the molecules in the bacterial enzymes and the milk, so a much smaller proportion of collisions have at least Ea, and the souring reactions slow down.
1.6 The nature of the reactants and the role of a catalyst
Different substances react at different rates under the same conditions, because they need different bonds broken and so have different activation energies:
- Reactive metals lose electrons more easily. In the same acid, magnesium fizzes violently, zinc steadily, iron slowly and copper not at all (May 2023 Chemistry asked why Cavendish’s metals gave hydrogen at different rates).
- Reactions between ions in solution, such as the precipitation Ag⁺(aq) + Cl⁻(aq) → AgCl(s), are almost instant: oppositely charged ions attract and no covalent bonds need to be broken.
- Reactions that break strong covalent bonds are slow at room temperature: rusting, the burning of methane without a spark, and the decay of plastics.
A catalyst speeds up a reaction without being used up, by providing an alternative pathway with a lower activation energy. Lesson 3 is devoted to catalysts; for now, remember that a catalyst changes the fraction of collisions that succeed, not the collision frequency.
A student writes: “Powdered magnesium reacts faster than ribbon because the powder particles have more energy. It also makes more hydrogen because there is more surface.” Find both errors and correct them.
Reveal answer
Error 1: the powder’s particles do not have more energy — the temperature is the same. The powder has a greater surface area, so acid particles collide with magnesium atoms more often; the frequency of collisions, and so of successful collisions, rises.
Error 2: the total volume of hydrogen depends only on the amount of magnesium (and acid), not on its surface area. Same mass of magnesium, same total hydrogen — it is simply produced faster.
Explain why reducing the volume of a container of two reacting gases, at constant temperature, increases the rate of reaction.
Reveal answer
The same number of molecules is in a smaller volume, so the pressure and the concentration (molecules per dm³) increase. The molecules move at the same average speed, so they collide more frequently, and with the same fraction of collisions having at least Ea there are more successful collisions per second.
A student says: “Temperature affects rate only because particles collide more often when they move faster.” At 20 °C the rate of a reaction is 1.0 unit; at 30 °C it is 2.0 units. Use the fact that the average speed of gas molecules is proportional to √T (T in kelvin) to show that faster collisions alone cannot explain the doubling.
Reveal answer
Speed ratio = √(303 ÷ 293)
Speed ratio = √1.034
Speed ratio = 1.017
Collision frequency is proportional to speed, so it rises by only about 1.7 %. The rate rose by 100 %, so almost all of the increase must come from the much larger fraction of collisions with energy at least equal to Ea. The student’s statement is incomplete — the energy effect dominates.
Try it — what speeds up marble and acid?
The lab reacts marble chips (calcium carbonate) with hydrochloric acid and collects the carbon dioxide in a gas syringe for exactly 60 s. The slider sets the acid temperature from 10 to 60 °C; you can also change the acid concentration (0.5–2.0 mol dm⁻³) and the chip size (large, medium, small chips or powder).
The syringe holds 100 cm³, so choose conditions that keep the volume below that. Pick ONE independent variable and keep the other two fixed.
Suppose your mean volumes in 60 s with 0.5 mol dm⁻³ acid and medium chips were: 10 °C → 3.1; 20 °C → 6.0; 30 °C → 11.8; 40 °C → 24.3; 50 °C → 47.6 cm³.
(a) Calculate the mean rate at 20 °C and at 50 °C in cm³ s⁻¹.
(b) Test the rule “the rate doubles for every 10 °C rise” using three pairs of results.
(c) Explain, with collision theory, why the graph of volume against temperature is a curve that gets steeper.
Reveal answer
(a) Rate at 20 °C = 6.0 cm³ ÷ 60 s
Rate at 20 °C = 0.10 cm³ s⁻¹
Rate at 50 °C = 47.6 cm³ ÷ 60 s
Rate at 50 °C = 0.79 cm³ s⁻¹
(b) 10 → 20 °C: 6.0 ÷ 3.1 = 1.9
20 → 30 °C: 11.8 ÷ 6.0 = 2.0
30 → 40 °C: 24.3 ÷ 11.8 = 2.1
Each 10 °C step multiplies the volume by about 2, so the rule holds within experimental scatter for this reaction.
(c) Each 10 °C rise multiplies the rate by the same factor (about 2), not adds the same amount, so the increases get bigger and bigger — an exponential-type curve. The reason is that a small rise in average energy pushes a large additional fraction of particles above Ea, and that fraction grows by a similar factor for each step.
Common mistakes to avoid
- “More energy” without “collisions” — every rate explanation must mention collisions: more frequent, or more with at least Ea, or both.
- Concentration gives the particles more energy — no: it gives more particles per unit volume, so more frequent collisions; the energy is set by the temperature.
- Rate and amount confused — a faster reaction does not make more product; the amount depends on the limiting reactant.
- Surface area “of the particles” — say “greater surface area of the solid, so more particles exposed to collisions”.
- Temperature explained by frequency alone — the main effect is the larger proportion of collisions with enough energy.
- “Rate” as the dependent variable — you measure a volume, a mass or a time; rate is calculated from it.
- Catalyst “gives particles energy” or “is used up” — it lowers Ea via another pathway and is chemically unchanged at the end.
Study smart: how this is examined
Study routine. Today: write the three conditions of collision theory and the five factors, each with its one-sentence particle reason, on one card. In two days: redo the cube and sugar-dust EXAMPLES from memory and the Stretch check on temperature.
In a week: run the marble lab for temperature and for chip size, then answer the two past-paper items above aloud in three linked points each before trying the quiz.
How we found out — measuring how fast
For most of history chemists asked whether substances react, not how fast. In 1850 the German chemist Ludwig Wilhelmy followed the breakdown of cane sugar in acid by measuring how the solution rotated polarised light, reading by reading, over hours. He found that the rate at each moment was proportional to the amount of sugar still left — the first quantitative rate law.
In Oxford in the 1860s Augustus Vernon Harcourt, a chemist, and William Esson, a mathematician, timed reactions such as hydrogen peroxide with hydrogen iodide and showed how the rate depended on the concentration of each reactant and on the temperature. Their careful graphs made “rate” a measurable quantity.
Why temperature mattered so much was explained in 1889 by the Swedish chemist Svante Arrhenius. He proposed that only molecules with more than a certain energy — what we now call the activation energy — can react, and that raising the temperature increases the number of such molecules very steeply.
Collision theory itself was put together independently by Max Trautz in Germany in 1916 and William Lewis in Britain in 1918. They used the kinetic theory of gases to count collisions per second and multiplied by the fraction with enough energy. For simple gas reactions the numbers came close to measured rates; for others they were too high, which is how the orientation condition entered the theory.
Where it is used now
- Food and cold chains. Refrigerators, cooler boxes and the cold rooms of abattoirs slow the reactions of spoilage bacteria and enzymes. Drying mopane worms, biltong or dried fruit removes water so particles in solution cannot meet.
- Dust and explosions. Grain silos, flour mills, sugar refineries and coal mines control dust, because a finely divided fuel in air has an enormous surface area. Coal-fired power stations deliberately pulverise coal so that it burns fast and completely.
- Mining and metals. Across southern Africa, copper and gold ores are crushed and ground to a fine powder before chemicals are added to extract the metal, partly so that the reagents reach more surface in less time.
- Cooking. A pressure cooker raises the boiling point of water to about 120 °C, so beans and tough meat cook in a fraction of the time. Chopping vegetables smaller does the same through surface area.
- Medicines. Tablets are designed to dissolve at a chosen rate: a fine powder acts quickly; a coated or slow-release tablet limits the surface exposed to stomach acid.
Science now — and the scientist you need to be
What is happening in 2026
Controlling rate is now a design problem that chemists share with engineers and computers. Pharmaceutical manufacturers increasingly run reactions in continuous-flow reactors — narrow tubes in which reactants meet at a precise temperature for a precise number of seconds — instead of in large batch vessels. Small channels give excellent mixing and heat removal, so fast, strongly exothermic reactions can be controlled safely.
Research groups are building automated laboratories in which robots run hundreds of experiments and machine-learning software chooses the next temperature, concentration or catalyst to try. The chemistry of collisions has not changed. What has changed is how quickly the best conditions can be found and how much data a scientist must be able to judge.
The scientist the next decade needs can look at a rate graph and say whether a change came from more collisions or from more energetic ones — and can check whether a computer’s suggestion makes chemical sense.
The mindshift
What you need to be able to do
Calculate a mean rate with its unit; explain every factor in one sentence that mentions collisions; separate frequency from energy for temperature; calculate a change in surface area; and keep rate and amount of product apart in every answer.
See an example
Typical times: whole tablet, cold water, about 90 s; crushed, cold, about 30 s; whole, warm, about 40 s. The crushed tablet has a much larger surface area, so water and acid particles collide with more of the solid each second. The warm water gives a larger proportion of collisions with energy at least Ea.
If the assistant says the crushed tablet “has more energy”, it has made the classic error.
Key terms
| Rate of reaction | The change in amount or concentration of a reactant or product per unit time; units such as cm³ s⁻¹, g s⁻¹, mol dm⁻³ s⁻¹. |
| Mean rate | The total change divided by the total time over an interval. |
| Collision theory | Particles react only when they collide with energy at least equal to the activation energy and with a suitable orientation. |
| Activation energy, Eₐ | The minimum energy that colliding particles must have for a reaction to occur. |
| Successful collision | A collision that leads to reaction: enough energy and the right orientation. |
| Collision frequency | The number of collisions per second; raised by higher concentration, pressure or surface area, and slightly by temperature. |
| Surface area | The area of a solid exposed to the other reactant; greater when the solid is broken into smaller pieces. |
| Concentration | The amount of a substance per unit volume of solution, in mol dm⁻³. |
| Catalyst | A substance that increases the rate of a reaction by providing an alternative pathway of lower activation energy, and is chemically unchanged at the end. |
| Limiting reactant | The reactant that is used up first and so fixes the total amount of product. |
Think like a scientist
Criterion B A statue company wants to know how quickly acid rain attacks the different sizes of marble chip it sells for garden paths. A school has marble chips sorted with sieves into five size ranges (0–1, 1–3, 3–6, 6–9 and 9–12 mm), 0.10 mol dm⁻³ hydrochloric acid, a balance reading to 0.01 g, conical flasks, cotton wool, a stopwatch, a thermometer and a 100 cm³ measuring cylinder.
The plan is to put the flask of acid and chips on the balance, plug it with cotton wool (which lets carbon dioxide out but stops acid spray), and follow the mass.
(a) Write a research question naming both variables with a range. [2]
(b) Formulate a hypothesis with a prediction and a “because” at the particle level. [3]
(c) Identify the independent and dependent variables with units and instruments, and explain why “rate” is not an acceptable dependent variable. [3]
(d) Identify three control variables, each with a value and a reason. [6]
(e) Write a numbered method including repeats, what is recorded and how the rate is processed, and one specific safety point. [6]
(f) Explain why the chips in each size range should have the same total mass, not the same number of chips. [2]
Model answer
(a) How does the size of marble chips (0–1 mm to 9–12 mm, five sieve ranges) affect the rate of mass loss (g s⁻¹) when 10.00 g of chips react with 50 cm³ of 0.10 mol dm⁻³ hydrochloric acid at 20 °C?
(b) If the chips are smaller, then the mass lost in the first 2 minutes will be greater — the smallest chips giving the fastest rate — because the same mass of smaller chips has a larger total surface area.
More calcium carbonate particles are exposed, so acid particles collide with the marble more often, giving more successful collisions per second and faster loss of carbon dioxide.
(c) IV: chip size range (mm), set by sieving. DV: mass of the flask and contents (g), read on the balance to 0.01 g every 30 s; the mass lost is calculated.
“Rate” is not measured directly: it is calculated from the mass lost divided by the time, so the measured DV is the mass (and the time).
(d) Mass of marble: 10.00 g in every run, because a larger mass has a larger total surface area and would react faster whatever its size.
Acid: 50 cm³ of 0.10 mol dm⁻³ from the same bottle, because concentration changes collision frequency.
Temperature: 20 °C, measured with the thermometer before each run, because higher temperature raises the fraction of collisions with at least Ea. (“Same equipment” is not a control variable.)
(e) 1. Weigh 10.00 g of 0–1 mm chips. 2. Measure 50 cm³ of acid into the flask, check it is at 20 °C, place it on the balance with the cotton wool beside it and tare. 3. Add the chips, plug with cotton wool, start the stopwatch.
4. Record the mass reading every 30 s for 5 minutes. 5. Repeat twice more (three trials), then repeat for each size range. 6. Calculate the mass lost at each time and the mean rate over the first 2 minutes (mass lost ÷ 120 s); plot mean rate against the middle of each size range.
Safety: hydrochloric acid is an irritant to eyes — wear goggles; the cotton wool stops acid spray escaping.
(f) Same total mass means the same amount of calcium carbonate, so the only difference between runs is how that mass is divided. The same number of chips would give the large chips far more mass and more surface in total, confusing size with amount.
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth. Take the molar volume of a gas as 24 000 cm³ mol⁻¹.
A reaction produces 72 cm³ of oxygen in 45 s. Calculate the mean rate in cm³ s⁻¹ and in mol s⁻¹.
Reveal answer
Rate = 72 cm³ ÷ 45 s
Rate = 1.6 cm³ s⁻¹ [1]
n = V ÷ 24 000 cm³ mol⁻¹
n = 72 cm³ ÷ 24 000 cm³ mol⁻¹
n = 3.0 × 10⁻³ mol
Rate = 3.0 × 10⁻³ mol ÷ 45 s
Rate = 6.7 × 10⁻⁵ mol s⁻¹ [1]
State the two conditions, other than colliding, that a collision must meet to be successful.
Reveal answer
Energy at least equal to the activation energy [1]; a suitable orientation of the particles [1].
Explain, using collision theory, why a higher concentration of hydrogen peroxide whitens teeth faster.
Reveal answer
At a higher concentration there are more hydrogen peroxide particles in the same volume [1]; so there are more frequent (successful) collisions with the stain molecules per second [1]; so the stains are broken down and the teeth whitened more quickly — the rate is faster [1].
Explain why powdered coal is burned in power stations rather than lumps.
Reveal answer
Powdered coal has a much greater surface area than the same mass of lumps [1]; so oxygen molecules collide with carbon at more sites per second and the rate of combustion is faster [1]; so energy is released more quickly (and the coal burns more completely) [1].
A reaction takes 240 s at 25 °C. Assuming the rate doubles for every 10 °C rise, estimate the time at 45 °C.
Reveal answer
Two 10 °C steps, so rate × 2² = × 4 [1]
Time = 240 s ÷ 4 = 60 s [1]
Explain why increasing the temperature increases the rate of a reaction. Your answer should include two separate effects.
Reveal answer
Particles have more kinetic energy and move faster, so they collide more frequently [1]; a greater proportion of collisions have energy equal to or greater than the activation energy [1]; so there are more successful collisions per second — the second effect is the larger one [1].
7. 1.20 g of magnesium ribbon (M = 24.3 g mol⁻¹) is added to 50.0 cm³ of 1.00 mol dm⁻³ hydrochloric acid: Mg(s) + 2HCl(aq) → MgCl₂(aq) + H₂(g). Determine the limiting reactant and the total volume of hydrogen. Then state and explain how the initial rate and the total volume would change if the same mass of magnesium powder were used with 100 cm³ of 0.50 mol dm⁻³ acid.
Reveal answer
n(Mg) = 1.20 ÷ 24.3 = 0.0494 mol
n(HCl) = 1.00 × 0.0500 = 0.0500 mol
HCl needed = 2 × 0.0494 = 0.0988 mol, more than is present, so HCl is limiting [1]
n(H₂) = 0.0500 ÷ 2 = 0.0250 mol
V = 0.0250 × 24 000 = 600 cm³ [1]
New run: the acid is still 0.0500 mol (0.50 × 0.100), so the total volume is still 600 cm³ [1].
The initial rate is affected two ways: the powder has a much greater surface area (faster), but the acid concentration is halved (fewer collisions per second per unit area); the powder effect is usually far larger, so the initial rate is expected to be greater — a prediction to test, because the two changes act in opposite directions [1].
8. A grain silo’s dust consists of particles about 0.05 mm across. A student claims: “Dust explodes because each grain particle has more energy than a whole grain.” Evaluate the claim and calculate how many times greater the surface area of 1.0 cm³ of grain is when it is ground into cubes of side 0.050 mm than when it is a single 1.0 cm cube.
Reveal answer
Side 0.050 mm = 0.0050 cm
Number of cubes = (1.0 ÷ 0.0050)³
Number of cubes = 8.0 × 10⁶ [1]
SA = 8.0 × 10⁶ × 6 × (0.0050 cm)²
SA = 1 200 cm²
Ratio = 1 200 ÷ 6.0 = 200 times greater [1]
The claim is wrong: the dust is at the same temperature as the grain, so its particles have the same energy [1]. The explosion happens because the enormous surface area, surrounded by air, allows oxygen to collide with fuel at a vast number of sites at once; the reaction is exothermic, so the energy released raises neighbouring particles above Ea almost instantly [1].
Summary
- Rate of reaction is the change in amount of reactant or product per unit time; it is calculated from measured changes, with units such as cm³ s⁻¹, g s⁻¹ or mol dm⁻³ s⁻¹.
- Collision theory: particles must collide, with energy at least equal to the activation energy and a suitable orientation. Rate = successful collisions per second.
- Concentration, gas pressure and surface area raise the collision frequency; the fraction that succeed is unchanged.
- Temperature raises the frequency a little and the fraction of collisions with at least Eₐ a lot — often doubling the rate per 10 °C.
- The nature of the reactants sets Eₐ; a catalyst provides a lower-Eₐ pathway. None of these factors changes the total amount of product.
Next time → Measuring rate: gas syringes, balances and the disappearing cross; reading rate graphs, and drawing tangents to find the rate at any instant.