Full chapter
Rate of Reactions
Explain why reactions speed up, and read what a progress graph actually proves.
O-Level 6092 (2026) / SEC G3 K324 (2027)
Explain a rate change using collisions
Changing a condition changes how often particles react.
Make a guess. You are not marked.
Marble chips react with excess dilute hydrochloric acid. The experiment is repeated with the same mass of marble as a fine powder. What changes?
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Faster reaction, same total volume of gas
Powder exposes more surface, so acid particles collide with it more often. The mass of marble is unchanged, so the total carbon dioxide is unchanged.
Reaction rate measures how quickly a reactant is used up or a product forms. Reacting particles must collide, and collisions must have enough energy to lead to reaction. An explanation should connect the changed condition to particle behaviour, then to the number of successful collisions per unit time.
| Change | Particle explanation | Important control |
|---|---|---|
| Higher solution concentration | More reacting particles per unit volume; more frequent collisions | Keep total solution volume and temperature controlled |
| Higher gas pressure at fixed temperature | Gas particles are closer together; more collisions per unit volume per second | Applies to gaseous reactants, not simply pressing on a solid |
| Smaller solid pieces, same total mass | Greater exposed surface area offers more places for collisions | Grinding does not create more total moles of solid |
| Higher temperature | Particles gain kinetic energy and move faster. A greater proportion of particles have energy equal to or greater than the activation energy (the minimum energy needed to react), so the frequency of effective collisions increases | Do not explain temperature solely as an increase in concentration |
A larger lump and equal-mass powder can produce the same final amount of gas while reacting at different speeds. In contrast, adding more of a limiting reactant can change the final amount. Keep the question "how fast?" separate from "how much?".
Check your understanding
Why does powdered calcium carbonate react faster than the same mass of chips in excess acid?
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The powder has more surface area, so there are more collisions with acid each second.
Acid particles can only hit the exposed surface. Powder exposes far more surface, so effective collisions are more frequent. The final volume of CO2 is the same, because the mass of carbonate is the same.
Exam-style question
Magnesium ribbon reacts with dilute hydrochloric acid. The experiment is repeated with more concentrated acid, keeping everything else the same. Explain, in terms of particles, why the reaction is faster. [2]
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Exam-style question
Calcium carbonate chips react with dilute hydrochloric acid. The experiment is repeated at a higher temperature, keeping everything else the same. Explain, in terms of particles, why the reaction is faster. [3]
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Choose a changing quantity
Match the measurement to the reaction.
| Measurement | Suitable situation | Rate evidence |
|---|---|---|
| Gas volume against time | Gas-producing reaction; gas can be collected | Volume increase per unit time |
| Mass against time | A gas escapes from an open reaction vessel | Mass loss per unit time |
| Time to a fixed visible endpoint | A precipitate obscures a mark or a colour reaches a chosen endpoint | Shorter time means a faster average approach to that same endpoint |
For a gas syringe, check airtight connections and free plunger movement. For mass loss, prevent splashes so that lost liquid is not mistaken for escaped gas. Start timing consistently. A subjective colour or visibility endpoint is less precise than a clear instrumental measurement, so use the same criterion and repeat.
Worked exampleCalculate an average rate
A reaction produces 36 cm3 of gas in its first 40 s.
Try it first. Then show the worked stepsHide the worked steps
- Average rate = change in gas volume / time interval.
- 36/40 = 0.90 cm3 s-1.
- This is an average over 40 s; the rate may have fallen throughout that interval.
0.90 cm3 s-1. State both interval and units.
Try it step by step
Work it out in steps: Change in the measured quantity, then Time interval, then Average rate.
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Problem 1.
Change in the measured quantity: Change in volume = 42.0 - 12.0 = 30.0 cm3.
Time interval: Time interval = 80 - 20 = 60 s.
Average rate: Average rate = 30.0 / 60 = 0.500 cm3 s-1 over that interval.
Problem 2.
Change in the measured quantity: Mass lost = 120.84 - 120.24 = 0.60 g.
Time interval: Time interval = 120 - 0 = 120 s.
Average rate: Average rate = 0.60 / 120 = 0.0050 g s-1. Two significant figures, because 0.60 g has two.
Problem 3.
Change in the measured quantity: Change in volume = 39.0 - 15.0 = 24.0 cm3.
Time interval: Time interval = 50 - 10 = 40 s.
Average rate: Average rate = 24.0 / 40 = 0.600 cm3 s-1.
Check your understanding
Two experiments reach the same end-point, one in 20 s and one in 40 s. Which is faster?
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The 20 s one, because the same change happened in less time.
For the same amount of change, rate is proportional to 1/time. 1/20 is twice 1/40, so the 20 s experiment is twice as fast.
Read slope and final amount separately
A plateau means product formation has stopped, not that nothing happened.
Make a guess. You are not marked.
Magnesium ribbon (in excess) reacts with 50 cm3 of dilute hydrochloric acid, and the gas volume is plotted against time. The experiment is repeated with 50 cm3 of acid at half the concentration. How does the new curve compare?
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Less steep, and it levels off at half the volume
Fewer acid particles per cm3 means fewer collisions per second. The acid is the limiting reactant and there are half as many moles of it, so half as much gas forms.
Same yield, different rates
The faster experiment reaches 44 cubic centimetres earlier; both finish at the same volume.
The gradient of a product-volume graph gives rate. A steeper initial slope means faster initial reaction. The curve becomes less steep as reactants are consumed. A horizontal line means no further gas is collected; a reactant may have been exhausted. A tangent estimates the instantaneous rate at a point; a line joining two measured points gives the average over that interval.
For a mass-loss graph, the mass decreases, so the slope is negative. Use the magnitude of the decrease per time for a positive reaction-rate value. Compare final gas volumes only at the same temperature and pressure. A leak can reduce apparent volume without changing the actual chemical yield.
Worked exampleDiagnose a changed curve
A repeat reaches its plateau sooner but at the same final volume. What can be concluded?
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- It formed the measured product more quickly.
- The same final amount formed under the stated gas conditions.
- A rate factor such as temperature, concentration or surface area may explain it; the curve alone does not identify which.
Faster rate, same final amount. Do not claim that a catalyst or temperature change is uniquely proved.
Check your understanding
A lower final volume of gas is recorded after a bung leaks. Does that prove less product formed?
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No. Some gas probably escaped before it could be collected.
A leak changes the measurement, not the chemistry. Separate apparatus faults from chemical explanations before drawing a conclusion.
Exam-style question
Magnesium ribbon reacts with excess dilute hydrochloric acid. The volume of hydrogen is recorded every 30 s.
Time / s: 0, 30, 60, 90, 120, 150, 180
Volume / cm3: 0, 20, 34, 43, 48, 50, 50
(a) Plot the results and draw a suitable line. [2]
(b) Use your graph to find the volume at 45 s. [1]
(c) State the time at which the reaction stopped, and explain why it stopped. [1]
(d) The experiment is repeated with the same mass of magnesium and acid that is twice as concentrated, still in excess. Sketch the new curve on the same axes and explain its shape. [2] [6]
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Check your understanding
Excess zinc reacts with 50 cm3 of dilute sulfuric acid, so the acid is limiting. It is repeated with 50 cm3 of acid of half the concentration. How does the new volume-time curve compare?
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It starts less steeply and levels off at half the volume
Fewer acid particles per unit volume means fewer effective collisions, so the starting gradient is lower. Half the concentration in the same volume is half the moles of the limiting acid, so half the hydrogen forms.
Catalysts provide an easier reaction pathway
They change the rate without being used up overall.
Make a guess. You are not marked.
A catalyst is added to a reaction that would otherwise be slow. What happens to the total amount of product made by the end?
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It stays the same; it just forms sooner
A catalyst makes the product form faster, but the amounts of reactants decide how much product can form.
A catalyst increases reaction rate and is chemically unchanged overall at the end. It can take part in intermediate steps before being regenerated. It provides an alternative pathway with lower activation energy, so a larger fraction of collisions can succeed at the same temperature. It does not supply the reaction energy or increase the final yield from a fixed limiting reactant.
| Context | Catalyst or catalytic material |
|---|---|
| Haber ammonia synthesis | Iron |
| Hydrogenation of unsaturated oils | Nickel |
| Hydrogen peroxide decomposition | Manganese(IV) oxide |
| Sulfuric-acid manufacture | Vanadium(V) oxide, an example of a compound acting as an industrial catalyst |
| Catalytic converters | Metal catalysts speed conversion of exhaust pollutants |
| Biological reactions | Enzymes are biological catalysts |
On an energy profile, the catalysed route has a lower peak but the same reactant and product levels. Enzymes also increase rates, but their activity depends on conditions; heating them too much can change their structure and reduce activity. Do not treat every catalyst as able to work under any conditions.
Check your understanding
Adding a catalyst gives more gas after 20 s. Will it give more gas once the reaction is complete?
Show the answer
No. It makes the same final amount of gas, just sooner.
A catalyst speeds up the reaction but does not change how much reactant there is. The final amount is set by the limiting reactant.
Exam-style question
Manganese(IV) oxide speeds up the decomposition of hydrogen peroxide, 2H2O2(aq) → 2H2O(l) + O2(g).
(a) Define the term catalyst. [1]
(b) Explain, in terms of activation energy and collisions, how manganese(IV) oxide increases the rate. [2] [3]
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Check your understanding
Which statement about enzymes is correct?
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They are biological catalysts that are denatured if heated too much
Enzymes are proteins made by living things that speed up reactions, such as yeast enzymes in fermentation. High temperatures change their shape, so they stop working.
Plan a fair rate investigation
A plan must say how the measurements will support a conclusion.
- Independent variable
Prepare a range of acid concentrations by measured dilution, keeping total solution volume equal.
- Controls
Use equal magnesium masses and exposed areas, the same initial temperature and the same apparatus.
- Measurements
Record hydrogen volume at fixed times after mixing; repeat each concentration.
- Analysis
Plot volume-time curves and compare initial gradients, with units. A repeat checks reproducibility.
- Practical precautions
Keep away from ignition sources because hydrogen is flammable; use suitable eye protection and avoid a sealed vessel without a free gas outlet.
To investigate particle size instead, vary chip/powder size while holding carbonate mass and acid concentration/volume constant. Use acid in excess if you want final gas amount fixed by carbonate. A suitable method controls the variables that could give the same observed effect, not just the easiest variables to measure.
Check your understanding
A student compares hot powdered carbonate with cold carbonate chips. Can the difference in rate be put down to temperature?
Show the answer
No. Particle size changed as well, so either factor could explain it.
Two variables changed at once. To test temperature, keep the particle size the same; to test particle size, keep the temperature the same.
Quick revision
Revisit the essentials, then return to an explanation when you need it.
| Term | Definition to write |
|---|---|
| Rate of reaction | The change in the amount (or concentration) of a reactant or product per unit time. |
| Effective collision | A collision between reacting particles that have energy equal to or greater than the activation energy, so that they react. |
| Term | Definition to write |
|---|---|
| Catalyst | A substance that increases the rate of a reaction and remains chemically unchanged at the end of the reaction. |
| Enzyme | A biological catalyst (a protein). It is denatured at high temperatures. |
Rate describes change per unit time. Concentration, pressure, temperature and exposed surface area affect successful collisions. On a progress graph, slope indicates rate and the final level indicates measured amount.
Pure: a catalyst lowers the activation barrier and is regenerated. A fair investigation changes one variable, controls alternatives and states how data will be analysed.
Scope
Learning outcomes
10. Rate of Reactions (6092 / K324). Use the outcome map to find the explanation for a particular syllabus requirement.
See the learning outcome map
10(a) Explain rate factors
- Concentration
- Gas pressure
- Particle size/surface area
- Temperature
- Collisions between reacting particles
10(b) Define catalysts and their effects
- Rate increase
- Chemically unchanged overall
- Enzymes included
10(c) Explain lower activation energy
- Alternative pathway
- More successful collisions
10(d) Recognise catalyst applications
- Industrial processes
- Enzymes as biological catalysts
10(e) Design rate investigations
- Choose variable and method
- Control confounders
- Measurements and analysis
10(f) Interpret rate data
- Volume/mass/time measurements
- Gradient and plateau
- Measurement limitations
Choose a changing quantityRead slope and final amount separately