OxfordAQA International GCSE Chemistry: Rate of reaction (9202) – Revision Notes

For full explanations and longer worked examples, use the rate of reaction study guide. Use these pages for quick recall once you have learned the content.

Their source is the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards. The notes summarise section 3.8.1 Rate of reaction, statements a to h, plus its required practical. With no tiers in this qualification, every line applies to you, and either paper can test this unit.

Links: practice questions · OxfordAQA chemistry hub · topic checklist · free diagnostics

3.8.1 a: what rate means and how to calculate it

Rate of reaction = how quickly reactants become products.

You measured	Formula	Unit
Mass of reactant used	mass used ÷ time	g/s
Volume of gas formed	volume formed ÷ time	cm³/s
Mass of product, converted to moles	moles ÷ time	mol/s
Time for a visible change	1 ÷ time	s⁻¹ (a relative rate)

Method in steps: mean rate from a table

1. Pick the two times that bound the interval asked for.
2. Subtract to find the amount used or formed in that interval.
3. Subtract to find the time taken.
4. Divide amount by time. Add the unit.

Worked reminder: rate from a time. At one concentration a precipitate hides the mark after 125 s; at another it takes 50 s. 1/time = 1 ÷ 125 = 0.008 s⁻¹ and 1 ÷ 50 = 0.02 s⁻¹. The second rate is 2.5 times the first, so that solution reacts faster.

Worked reminder. A gas syringe reads 21 cm³ at 20 s and 57 cm³ at 50 s. Mean rate = (57 − 21) ÷ (50 − 20) = 36 ÷ 30 = 1.2 cm³/s.

Reading the graph

Feature of an amount-of-product curve	What it tells you
Steepest part at the start	Rate is highest; reactants most concentrated
Gradient getting smaller	Reaction slowing; reactant particles being used up
Horizontal line	Reaction has stopped; a reactant is used up
Height of the flat section	Total amount of product
Steeper of two curves	The faster reaction
Two curves, same final height	Same amounts of reactants; only the rate differs
One curve levels off lower	Less limiting reactant was used

A reactant curve is the mirror image: it falls steeply, then levels off.

Method in steps: sketching a second curve on the same axes

1. Decide whether the change alters the rate, the final amount, or both.
2. Faster reaction (hotter, more concentrated, powdered, catalysed): start steeper and level off sooner.
3. Same limiting reactant: level off at the same height as the original curve.
4. Half the limiting reactant: level off at half the height.
5. Slower reaction: start less steeply and take longer to level off.

Worked reminder. A run makes 80 cm³ of gas. It is repeated with half the mass of the solid, which is the limiting reactant, ground to a powder. The new curve starts steeper but levels off at 40 cm³.

3.8.1 b: collision theory

• Particles react only when they collide with sufficient energy.
• Activation energy = the minimum energy particles need for a reaction to happen.
• Faster rate = more successful collisions per second.

3.8.1 c to f: the four factors

Factor increased	Effect on particles	Collision effect
Temperature	Particles move faster	More frequent and more energetic collisions
Pressure (gases)	More particles in the same volume	More frequent collisions
Concentration (solutions)	More particles in the same volume	More frequent collisions
Surface area (solids)	More particles exposed	More frequent collisions

Must-know distinction. Only temperature changes collision energy. The other three change collision frequency only.

Model sentence for temperature. “At a higher temperature the particles move faster, so they collide more frequently and more energetically. A greater proportion of collisions have at least the activation energy, so the rate increases.”

Model sentence for concentration. “A stronger solution packs more reactant particles into each cm³, so they collide more often and the rate rises.”

Surface area check. A 4 cm × 4 cm × 4 cm block has surface area 6 × 16 = 96 cm². Cut into sixty-four 1 cm cubes, it has 64 × 6 = 384 cm². Same volume, four times the surface.

Required practical: factors affecting rate

Method in steps: planning

1. Choose one independent variable (for example, acid concentration).
2. Decide how you will judge rate: gas volume at fixed times, mass loss, or time for a cross to vanish.
3. List control variables: volumes, masses, particle size, temperature, same apparatus.
4. Carry out repeats and take a mean, excluding anomalous results.
5. Plot a graph and describe the pattern.
6. Evaluate: sources of error and how to reduce them.

Method	Suits	Watch out for
Gas syringe or upturned cylinder	Reactions giving a gas	Gas lost before the bung goes in
Balance, open flask	Reactions giving a dense gas such as CO₂	Tiny mass changes with hydrogen; acid spray
Disappearing cross	Reactions forming a precipitate	Judging the end point by eye

Evaluating results.

• Anomalous result: a value that does not fit the pattern. Repeat that run and leave it out of the mean.
• Repeatable: you get similar results when you repeat the run yourself.
• Reproducible: someone else, or a different method, gets similar results.
• Random error (scatter, judging the cross) is reduced by repeats and a mean.
• Systematic error (gas lost each time before the bung goes in) shifts every result the same way; repeats do not remove it.

Safety points. Eye protection with acids. No flames near hydrogen. Ventilate if sulfur dioxide forms, as in the thiosulfate reaction.

3.8.1 g and h: catalysts

• A catalyst alters how fast a reaction goes and is not used up. Same mass at the end.
• Different reactions need different catalysts.
• A catalyst does not change the amount of product formed.
• Industry uses catalysts to reduce costs: faster production, lower temperatures and pressures (less energy), and the catalyst can be reused.

Specification example	Catalyst
Haber process	iron
Contact process	vanadium(V) oxide
Ethene + steam	phosphoric acid
Fermentation	enzymes in yeast

How catalysts lower the activation energy is in the energy changes unit (3.9.2 h).

Must-know distinctions

• Faster vs more: a faster reaction reaches the same final amount sooner. More product needs more limiting reactant.
• Frequency vs energy: concentration, pressure and surface area change frequency; temperature changes both.
• Mean rate vs rate at the start: a mean over the whole reaction is lower than the initial rate.
• Time vs rate: a longer time for the cross to vanish means a lower rate.

Quick self-test

1. Define activation energy.
2. 66 cm³ of oxygen forms in 30 s. Calculate the mean rate.
3. 0.39 g of zinc is used up in 150 s. Calculate the mean rate in g/s.
4. The cross disappears after 64 s. Calculate 1/time to 2 significant figures.
5. Find the mean rate, in mol/s, when 0.132 g of CO₂ (Mr = 44) escapes in 60 s.
6. Which factor increases the energy of collisions as well as their frequency?
7. Explain why raising the pressure speeds up a reaction between two gases.
8. Give two features of a catalyst.
9. A gas-volume curve becomes horizontal after 70 s. What does this mean?
10. Give two reasons why catalysts lower costs in industry.
11. Name the catalyst used in the contact process.
12. A balance reading falls by 0.48 g in 160 s. Calculate the mean rate in g/s.

Answers

1. The minimum energy particles need for a reaction to happen.
2. 66 ÷ 30 = 2.2 cm³/s.
3. 0.39 ÷ 150 = 0.0026 g/s.
4. 1 ÷ 64 = 0.015625 = 0.016 s⁻¹.
5. 0.132 ÷ 44 = 0.003 mol; 0.003 ÷ 60 = 5.0 × 10⁻⁵ mol/s.
6. Temperature.
7. The gas particles are closer together in a smaller volume, so they collide more frequently.
8. It changes the rate of reaction; it is not used up (it can be reused).
9. The reaction has stopped because one reactant has been used up.
10. Any two: faster production; lower temperature or pressure needed, so less energy; catalyst reused.
11. Vanadium(V) oxide, V₂O₅.
12. 0.48 ÷ 160 = 0.003 g/s.

Where marks are usually lost

• Missing the “more energetic” point for temperature.
• Using “more collisions” where “more frequent collisions” is needed.
• Saying higher temperature gives particles more activation energy. The activation energy stays the same; more particles reach it.
• Leaving off the unit, or giving cm³/s for a mass-loss experiment.
• Reading the finishing time as the last time in the table, not the first time the reading stops changing.
• Claiming a catalyst increases the amount of product.
• Saying a catalyst is “not involved” in the reaction. It takes part but is not used up.
• In planning answers, giving “the amount of acid” as a control variable without saying volume and concentration.
• Forgetting that a longer time to a visible change means a slower rate.

Official syllabus

OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards (OxfordAQA). Covered here: section 3.8.1 Rate of reaction, with its required practical.