OxfordAQA International GCSE Chemistry: Rate of reaction (9202)

Use this guide to learn rate of reaction from scratch for OxfordAQA International GCSE Chemistry (9202). Its source is the OxfordAQA International GCSE Chemistry (9202) specification, Version 6.2, for exams May/June 2018 onwards. Section 3.8.1 is taught in full, statements a to h, together with the required practical linked to it. There are no tiers in this qualification, so nothing here is optional. Either Paper 1 or Paper 2 may test any part of the content, this unit included.

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Coverage of section 3.8.1

Statement	Skill you need
3.8.1 a	Find a rate from the amount of reactant used, or product formed, over time; interpret graphs of amount against time
3.8.1 b	State that particles must collide with sufficient energy to react; define activation energy
3.8.1 c	Explain why a higher temperature increases the rate
3.8.1 d	Explain why a higher pressure of reacting gases increases the rate
3.8.1 e	Explain why a higher concentration of reactants in solution increases the rate
3.8.1 f	Explain why a larger surface area of a solid reactant increases the rate
Required practical	Investigate factors affecting the rate of a reaction
3.8.1 g	State that catalysts change the rate but are not used up, and that different reactions need different catalysts
3.8.1 h	Explain why catalysts matter in industrial processes: they increase rates and reduce costs

The specification says you do not need to know specific reactions beyond those in the subject content, so questions will give you the reaction. You bring the method: measuring a rate, reading the graph and explaining each factor.

3.8.1 a: measuring the rate of a reaction

The rate of a reaction tells you how quickly reactants are turned into products. You find it by measuring how much reactant is used up, or how much product forms, in a known time:

Rate from a reactant	Rate from a product
rate = amount of reactant used ÷ time	rate = amount of product formed ÷ time

The unit depends on what you measured. A gas volume in cm³ over seconds gives cm³/s (also written cm³ s⁻¹). A mass in grams gives g/s. If you convert the mass to moles first, the rate is in mol/s.

Three common ways to follow a reaction

• Collect a gas in a gas syringe, or an upturned measuring cylinder full of water, and read the volume at regular times.
• Measure the loss of mass as a gas escapes from an open flask on a balance. A cotton wool plug stops acid spray.
• Time a visible change. If a reaction makes a precipitate, time how long it takes for a mark under the flask to vanish from view. A longer time means a slower reaction, so 1/time is used as a measure of rate.

Worked example 1: rate from a gas-volume table

Zinc granules react with dilute sulfuric acid: Zn + H₂SO₄ → ZnSO₄ + H₂. The hydrogen is collected in a gas syringe.

Time (s)	0	15	30	45	60	75	90	105	120
Volume of H₂ (cm³)	0	27	48	63	73	79	82	84	84

(i) Mean rate over the first 30 s. rate = 48 ÷ 30 = 1.6 cm³/s

(ii) Mean rate between 60 s and 90 s. Volume formed in this interval = 82 − 73 = 9 cm³. rate = 9 ÷ 30 = 0.3 cm³/s

(iii) Mean rate for the whole reaction. The volume stops rising at 105 s, so the reaction took 105 s. rate = 84 ÷ 105 = 0.8 cm³/s

The answer to (ii) is much smaller than (i): the reaction slows as it goes.

Reading a rate graph

Plot volume (or mass of product) on the y-axis and time on the x-axis. Then read the shape:

• Steep at the start. The rate is highest at the beginning, when the reactants are most concentrated.
• Gradient falls. As reactants are used up, there are fewer reacting particles, collisions become less frequent and the curve flattens.
• Flat line. The reaction has stopped because one reactant has been used up. The height of the flat section is the total amount of product.
• Comparing two curves. The steeper curve shows the faster reaction. If two runs use the same amounts of reactants, they finish at the same height even if one is faster. A curve that levels off lower used less of the limiting reactant.

A graph of mass of reactant against time starts high, falls steeply, then levels off.

Worked example 2: rate from a mass loss, in g/s and mol/s

Calcium carbonate powder reacts with dilute nitric acid in an open flask on a balance: CaCO₃ + 2HNO₃ → Ca(NO₃)₂ + H₂O + CO₂

The reading falls from 148.25 g to 147.59 g in the first 120 s.

Mass of CO₂ lost = 148.25 − 147.59 = 0.66 g Mean rate = 0.66 ÷ 120 = 0.0055 g/s

To give the rate in mol/s, convert the mass of CO₂ to moles (Mr of CO₂ = 12 + 16 + 16 = 44): moles of CO₂ = 0.66 ÷ 44 = 0.015 mol Mean rate = 0.015 ÷ 120 = 1.25 × 10⁻⁴ mol/s

3.8.1 b: collision theory and activation energy

A reaction can only happen when reacting particles collide with each other. Not every collision works. The particles must collide with sufficient energy. The activation energy is the minimum energy that particles need for a reaction to happen.

So there are two ways to make a reaction faster:

1. Make collisions happen more often (a higher frequency of collisions).
2. Make more of the collisions carry at least the activation energy (more energetic collisions).

Every factor below works through one or both of these.

3.8.1 c: temperature

Raising the temperature makes the reacting particles move faster. Two things follow:

• They collide more frequently.
• They collide more energetically, so a greater proportion of collisions have at least the activation energy.

Both effects increase the rate. Temperature is the only factor here that changes the energy of collisions as well as their frequency.

3.8.1 d: pressure of reacting gases

For a reaction between gases, a higher pressure squeezes the same number of gas particles into a smaller volume. The particles are closer together, so they collide more frequently and the rate increases. Pressure does not change the energy of each collision.

3.8.1 e: concentration of solutions

A more concentrated solution has more reactant particles in the same volume. Collisions with the other reactant become more frequent, so the rate increases. Again, the energy of each collision is unchanged.

3.8.1 f: surface area of solids

A solid can only react at its surface. Cutting or grinding it into smaller pieces exposes more particles to the other reactant. This increases the frequency of collisions, so the rate increases. A powder reacts faster than the same mass of lumps.

You may be asked to compare surface areas with simple geometry. Total surface area of a cuboid = 2 × (lw + lh + wh). Cutting a solid into pieces keeps the volume the same but adds new faces, so the total surface area rises.

Required practical: investigate factors affecting the rate of a reaction

The table in section 6.2 of the specification ticks every skill for this practical: designing the procedure, controlling variables, assessing risk, collecting data, analysing it, drawing conclusions and evaluating. Be ready for questions on each.

Planning

• Independent variable: the factor you change, for example the concentration of the acid or the temperature.
• Dependent variable: what you measure to judge rate, such as the gas volume after a fixed time or the time for a mark to disappear.
• Control variables: everything else, kept the same. For a gas-collection experiment with acid and a metal, that means the volume of acid, the mass and particle size of the metal, the temperature and the apparatus.

Method using a change you can see

When sodium thiosulfate solution and dilute hydrochloric acid are mixed, the liquid slowly turns cloudy as solid sulfur forms. Put a flask on a paper with a dark cross drawn on it, add a fixed volume of thiosulfate solution, then add a fixed volume of acid and start the timer. Stop it when you can no longer see the cross from above. To study temperature, warm both solutions separately in a water bath to the target temperature before mixing, and check it with a thermometer.

Worked example 3: turning times into rates

Temperature (°C)	20	30	40	50
Time for cross to disappear (s)	150	82	44	25
1/time (s⁻¹)	0.0067	0.0122	0.0227	0.040

Working for the first column: 1 ÷ 150 = 0.0067 s⁻¹. The 1/time values rise as the temperature rises, so the rate increases with temperature. Plot 1/time against temperature and draw a smooth curve of best fit.

Risks and evaluation

• The thiosulfate reaction gives off sulfur dioxide, an irritant gas, so use small volumes and keep the room well ventilated.
• Wear eye protection with acids. Hydrogen is flammable, so keep flames away from a gas-collection experiment.
• Repeat each run and calculate a mean to improve repeatability. Leave out any anomalous result when you do this, and repeat that run.
• Judging when a cross “disappears” depends on the observer. A light sensor and data logger removes the judgement.

3.8.1 g: catalysts

A catalyst changes the rate of a chemical reaction but is not used up during the reaction. At the end, the same mass of catalyst is still there and can be used again. You can show this by filtering it off, drying it and weighing it.

Different reactions need different catalysts. A substance that speeds up one reaction may do nothing for another. Examples from this specification:

Process	Catalyst
Haber process (ammonia)	iron
Contact process (sulfur dioxide to sulfur trioxide)	vanadium(V) oxide, V₂O₅
Ethene + steam → ethanol	phosphoric acid
Fermentation of sugar to ethanol	enzymes in yeast
Cracking hydrocarbons	a hot catalyst

Transition metals and their compounds are often useful as catalysts. How a catalyst works, by giving a pathway with a lower activation energy, is taught in the energy changes unit (statement 3.9.2 h).

A catalyst does not change the amount of product. On a gas-volume graph, a catalysed run is steeper at the start but levels off at the same final volume.

3.8.1 h: catalysts in industry

Catalysts help industry in several ways:

• A faster rate means more product per hour from the same plant.
• An acceptable rate can be reached at a lower temperature or pressure, which saves energy and lowers the cost of equipment.
• The catalyst comes out of the reaction unchanged, so one charge of it lasts a long time.

Together these reduce costs.

Common errors

• Saying “more collisions” without “more frequent”, or explaining temperature with frequency only.
• Saying a catalyst “gives particles energy” or is used up.
• Reading the end of a reaction from the last time in the table rather than the first time the volume stops changing.
• Mixing units: g/s for masses, cm³/s for gas volumes, mol/s only after converting to moles.
• Saying a faster reaction makes more product. Only more reactant does that.

Next steps

• Condense this unit with the rate of reaction revision notes.
• Test yourself with the rate of reaction practice questions.
• Revise moles with the moles and quantitative chemistry guide.
• See what still needs work with a free 10-minute diagnostic.
• Browse every unit from the OxfordAQA chemistry hub, and track progress on the topic checklist.

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 from section 6.2.