3bPhysical Chemistry

Rates of reaction — Edexcel Linear - OL (Chemistry) Revision Notes & Topical Questions

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Lesson summary

Rates of reaction Edexcel O Level Linear Chemistry notes: describe experiments to investigate the effects of changes in surface area of a solid, concentration of a solution, temperature and the use of a catalyst on the rate of a reactionTemperature mainly increases rate by increa…

Rates of reaction Edexcel O Level Linear Chemistry: what to do next
Rate of reaction
Collision theory
Activation energy
Boltzmann distribution
Homogeneous catalyst

Section 1 of 12Process & sequence

Particles collide
E ≥ Ea?
Correct orientation?
Successful reaction

Measure gas volume or mass loss at intervals.

Plot against time.

Draw a tangent at the chosen time.

Rate = gradient of the tangent.

Section 2 of 12Keep going

Rates of reaction Edexcel O Level Linear Chemistry: what to do next

These Rates of reaction Edexcel O Level Linear Chemistry notes covered the key ideas, worked examples and exam technique for this topic. Check your Rates of reaction Edexcel O Level Linear Chemistry understanding with the practice questions below, then reread any Rates of reaction Edexcel O Level Linear Chemistry point you were unsure about before moving on.

Practice: Rates of reaction Edexcel O Level Linear Chemistry practice questions

Previous lesson: 3a Energetics — Notes

Next lesson: 3c Reversible reactions and equilibria — Notes

Course: Edexcel Linear – OL (chemistry) — all units

Official syllabus: Pearson Edexcel International GCSEs

Section 3 of 12Graph calculations

On a graph of volume against time, mean rate = total volume ÷ total time for a whole reaction, or rate over an interval = Δvolume ÷ Δtime.

Example data: 0 s, 0 cm³; 30 s, 24 cm³; 60 s, 38 cm³; 90 s, 46 cm³; 120 s, 50 cm³; 150 s, 50 cm³.

Rate 0–30 s = 24 ÷ 30 = 0.80 cm³/s; 30–60 s = 14 ÷ 30 = 0.47 cm³/s; 90–120 s = 4 ÷ 30 = 0.13 cm³/s.

The rate falls because the reactants are used up and their concentration decreases.

The reaction ends at 120 s.

Calculating from a mass graph

if a flask loses 0.88 g of CO₂ (M_r 44) the moles of CO₂ are 0.020 mol, which is equal to the moles of CaCO₃ that reacted (1:1), i.e. 2.0 g of marble.
Use the mass lost to calculate how much reactant was present.

Section 4 of 12Applications: safety and industry

Rates matter in daily life and industry.

Coal mines and flour mills have explosion risks because fine dust has a large surface area.

Food is kept in a freezer or fridge, and food packaging removes oxygen, to slow decay.

Industrial reactors use catalysts to allow lower temperatures; chemical plants raise the pressure of gas reactions; powdered solids are used to speed up reactions;

and the temperature is chosen to balance rate against energy cost and, for reversible reactions, yield.

In car engines the catalytic converter works best at high surface area, using a honeycomb of platinum-coated ceramic.

Section 5 of 12Rate and equilibrium together: choosing conditions

In industry the chosen conditions are always a balance of rate, yield and cost.

Rate and equilibrium together: choosing conditions

For irreversible reactions, only rate and cost matter: higher temperature, powdered solids and catalysts are used.

For reversible exothermic reactions a high temperature raises the rate but lowers the equilibrium yield,

so a moderate temperature and a catalyst are used; recycling unreacted material raises the overall yield.

For the decomposition of limestone to quicklime,

which is endothermic and irreversible if carbon dioxide escapes, a high temperature (about 1000 °C) and crushed limestone are used because the rate and the yield both benefit.

Type of reaction	Temperature choice	Other measures
Irreversible, exothermic	As high as is economic and safe	Catalyst, fine particles
Reversible, exothermic	Moderate compromise	Catalyst, pressure, recycle
Reversible, endothermic	High (helps both)	Remove product as it forms

Section 6 of 12Exam technique for rate questions

When asked to "describe" a graph, quote values: "the volume rises quickly at first, 24 cm³ in the first 30 s, then more slowly until it levels off at 50 cm³ after 120 s." When asked to "explain", give the reason in particle terms. When asked to sketch a new curve, keep the same final volume if the limiting reactant is unchanged, and make it steeper for a faster reaction.

Section 7 of 12Compare

Factors affecting rate
Factor	Change	Effect on rate	Explanation	Boltzmann effect	Example
Concentration	Increase	Faster	More frequent collisions	No change in shape	More acid → faster fizz
Pressure (gases)	Increase	Faster	More particles per volume	No change	Haber process
Temperature	Increase 10 °C	Roughly doubles	More particles have E ≥ Ea, more collisions	Curve flattens, shifts right	Food cooks faster
Surface area	Powder	Faster	More exposed particles	—	Powdered CaCO₃
Catalyst	Add	Faster	Lower Ea	More particles above new Ea	Fe in Haber
Light	Increase (some reactions)	Faster	Energy breaks bonds	—	Photochemical chlorination

Section 8 of 12Worked examples

From the data above, calculate the mean rate for the whole reaction. [2]2 marks
Why does the rate decrease as the reaction proceeds? [2]2 marks
0.88 g of carbon dioxide is lost from a flask. Calculate the mass of calcium carbonate that reacted. [3]3 marks
Explain why a catalytic converter uses a honeycomb structure. [2]2 marks

Section 9 of 12Key terms

Term	Meaning
Mean rate	Total change divided by total time
Collision frequency	Collisions per second
Surface area	Exposed area available for reaction
Catalytic converter	Device using catalysts to remove pollutants
Honeycomb	Structure with large surface area
Concentration	Amount per volume

Section 10 of 12Syllabus checklist

describe experiments to investigate the effects of changes in surface area of a solid, concentration of a solution, temperature and the use of a catalyst on the rate of a reaction

describe the effects of changes in surface area of a solid, concentration of a solution, pressure of a gas, temperature and the use of a catalyst on the rate of a reaction

explain the effects of changes in surface area of a solid, concentration of a solution, pressure of a gas and temperature on the rate of a reaction in terms of particle collision theory

know that a catalyst is a substance that increases the rate of a reaction, but is chemically unchanged at the end of the reaction

know that a catalyst works by providing an alternative pathway with lower activation energy practical: investigate the effect of changing the surface area of marble chips and of changing the concentration of hydrochloric acid on the rate of reaction between marble chips and dilute hydrochloric acid

practical: investigate the effect of different solids on the catalytic decomposition of hydrogen peroxide solution

Section 11 of 12Command terms

Command terms

Give / State / Name

Recall a fact; one idea per mark.

Define

Give the exact meaning using the specification keywords.

Describe

Give an account of what happens / the pattern in data, quoting values; no reasons needed.

Explain

Give reasons with linked statements (cause → effect). Each linked step is usually a separate marking point.

Suggest

Use knowledge in a new context; more than one answer may be accepted.

Compare

Similarities and differences; each point must refer to both items.

Calculate

Show working; include units; check significant figures. Method marks are given even if the final answer is wrong.

Discuss

Explore the issue from different angles and link the points.

Evaluate / Assess

Weigh strengths and weaknesses and reach a supported judgement.

Deduce

Draw a conclusion from the information given.

Justify

Support a statement/choice with evidence.

Section 12 of 12Exam technique

Saying rate is constant throughout a reaction.

Using total volume in the denominator of an interval rate.

Linking temperature only to collision frequency.

Reading the time at which the line is steepest as the end.

Last updated: 14 October 2026