Rates of reactionIB MYP Sciences: Revision notes
Section 1
Measuring the rate of reaction
The rate of reaction is how fast reactants are used up or products are made.
rate = amount of reactant used or product made ÷ time taken
Three common ways to measure it:
- Volume of gas collected in a gas syringe at regular times
- Mass change on a balance, when a gas escapes
- Time for a colour change or for a cross to disappear as a precipitate forms
Worked example: 60 cm³ of gas collected in 20 s gives a mean rate of 60 ÷ 20 = 3 cm³/s.
A shorter time means a faster reaction, so time and rate go in opposite directions.
Section 2
Factors affecting rate
Four things change the rate:
- Temperature: a higher temperature increases the rate.
- Concentration: a higher concentration of a solution, or higher pressure for gases, increases the rate.
- Surface area: smaller pieces, such as powder instead of lumps, increase the rate.
- Catalyst: a catalyst speeds up a reaction and is not used up.
Section 3
Collision theory
Particles must collide with enough energy to react. This minimum energy is the activation energy.
- Higher temperature: particles move faster, so they collide more often and more collisions have enough energy.
- Higher concentration: more particles in each cm³, so more frequent collisions.
- Larger surface area: more particles exposed, so more frequent collisions.
- Catalyst: provides a different pathway with a lower activation energy, so more collisions are successful.
Do not say that particles get bigger or that more particles appear when heated. Say they move faster and have more energy.
Section 4
Interpreting rate graphs
On a graph of product against time:
- The steeper the line, the faster the rate.
- The line is steepest at the start, when the reactants are at their highest concentration.
- The line flattens as reactants are used up.
- When the line is horizontal, the reaction has finished.
A faster reaction gives a steeper line that levels off sooner. If the same amounts are used, the final volume is the same.
Section 5
Designing a fair investigation
In a fair test you change one variable and keep the others the same.
- Independent variable: the one you change, such as concentration.
- Dependent variable: the one you measure, such as time or gas volume.
- Control variables: everything else, such as temperature and volume.
Repeat the experiment and calculate a mean to improve reliability. To evaluate, comment on sources of error. A cross that fades is subjective, so a light sensor could improve it. Include a safety precaution, such as eye protection.
Must Know
- Rate = amount ÷ time. Measure by gas volume, mass change or time for a colour change.
- Temperature, concentration, surface area and catalysts increase the rate.
- Collision theory: more frequent or more energetic collisions give a faster rate.
- Steeper line means faster rate. Flat line means the reaction has finished.
That's the notes covered.
Carry on to the next subtopic.
Exam questions on Rates of reaction
- A student in Auckland adds magnesium ribbon to dilute hydrochloric acid in a flask and collects the hydrogen gas in a gas syringe, recording the volume every 10 seconds. She plots the volume of gas against time. The line is steepest at the start and becomes horizontal after about 120 seconds.Explain why the reaction is fastest at the start.2 marks
- A student in Dublin drops one effervescent vitamin tablet into 200 cm³ of water at 10 °C and times how long it takes for the fizzing to stop. It takes 120 s. She repeats the experiment with water at 40 °C and it takes 35 s. The tablets are the same size.The student repeats the experiment at 10 °C using a tablet that has been crushed into a powder. Predict how the time for the fizzing to stop will change and explain your prediction.2 marks
- A student in Cape Town investigates how the concentration of sodium thiosulfate solution affects its reaction with hydrochloric acid, which makes a cloudy yellow precipitate of sulfur. She stands the flask on a cross drawn on paper and times how long it takes for the cross to disappear when she looks down through the liquid. For sodium thiosulfate solutions of concentration 0.10, 0.05 and 0.025 mol dm⁻³, the times are 40 s, 80 s and 160 s.Describe the pattern in the results and explain it using collision theory.3 marks
Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).