Planning a fair rate investigation
Key idea: Change one rate factor and control the others. Measure gas volume at regular times or time the same visible end-point, then repeat to check how much the results vary.
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Guided lessons · coming soon
Guided lesson
Guided steps
- Predict the starting rate and final gas volume for small and large chips.
- Read what you change and what you keep the same.
- Choose what to measure and what to keep the same.
- Read how to measure temperature and time the cross.
- Write the plan, describe the trend and explain it with collisions.
- Find the mistake, explain the extra gas and fix the comparison.
- Explain why the sunny window spoils the test, then fix the method.
- Work out the temperature to record, and where to measure it.
- Explain the cooling on mixing and improve the method.
- Compare the two rates and decide what the times show.
- Explain why the different depths matter, then write a fair plan.
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ContinueA fair rate investigation changes one factor and keeps every other factor that affects the rate the same. Only then can a difference in the results be put down to the factor that was changed.
Predict rate and final product amount
Check your understanding 1 Predict before planning
Two runs use equal masses of marble, one as small chips and the other as large chips. The acid is in excess, with the same concentration, volume and temperature. Predict the initial rate and final gas amount for the small chips compared with the large chips.
Show answer
Change one factor and keep the others controlled
Change one factor. Keep other quantities and conditions constant where the chosen factor allows it. For example, changing concentration while keeping solution volume fixed also changes the amount of acid. If you want to compare rates with the same final gas amount, keep the acid in excess in every run and use the same amount of the solid reactant.
For equal masses of marble reacting completely in excess acid, smaller chips expose more surface and react faster initially, but the final gas amounts match. Changing marble mass as well as chip size changes both the exposed surface and the amount available to react.
Measure the same quantity at regular times, or time the same visible end-point. Repeat each condition, investigate unusual results and calculate a mean from valid results. A result should not be excluded just because it differs from the others.
Guided practice 1
Plan a concentration comparison
Problem
A student investigates how hydrochloric-acid concentration affects its reaction with equal masses of marble chips. The acid is in excess in every run. Choose a suitable measurement and controls for marble mass, particle size and temperature.
Change concentration only
Hints
Hint 1
product
The reaction produces carbon dioxide.
Hint 2
fair test
Keep every rate factor except concentration unchanged.
Show solution step by step
Choose the evidence
Method
Record carbon-dioxide volume at fixed time intervals using a gas syringe.
Reason
Plot the readings as a volume–time graph: its gradient is the rate.
Working
Answer
gas volume / cm³ against time / sControl the comparison
Method
Keep marble mass, marble surface area, acid volume and temperature constant.
Reason
Each could otherwise change rate or final gas amount.
Working
Answer
Change concentration only. Use equal marble masses from the same chip-size range, measure equal acid volumes and bring the solutions to the same measured temperature. Check that the gas syringe is airtight and record at the same time intervals.
Quick check
A higher concentration may produce gas faster without increasing the final volume, because the same marble amount runs out. Final gas volume alone therefore cannot measure rate. Use equal masses from the same chip-size range to control surface area.
Control the reaction temperature
A water-bath setting is not a reading of the reacting mixture. If one solution is warm and the other is cool, mixing them changes the temperature. Bring both solutions to the intended temperature before mixing, then measure the actual mixture and check for drift during the reaction. Keep the flask, cross, viewing position, liquid depth, mixing and timing start consistent so each disappearance time represents the same cloudiness.
When a mixture cools approximately steadily, the mean of its starting and final temperatures is an estimate of its average temperature. If the temperature changes sharply or irregularly, two readings are not enough: record it during the run and improve temperature control.
Guided practice 2
From plan to data to conclusion
Problem
A student investigates how temperature affects the reaction between sodium thiosulfate solution and dilute hydrochloric acid, timing how long a cross under the flask takes to disappear. (a) State what to change, what to measure and three things to keep the same. (b) The results are: 20 °C, 120 s; 30 °C, 62 s; 40 °C, 33 s; 50 °C, 18 s. Describe the trend in rate. (c) Write a conclusion the data support and explain it with collisions.
Plan, read, conclude
Hints
Hint 1
time and rate
A shorter time to the same end-point means a faster rate.
Hint 2
scope
Only claim what happened between the temperatures you tested.
Show solution step by step
Plan
Method
Change only the temperature; measure the time for the cross to disappear.
Reason
Every other rate factor must stay fixed for the times to be compared.
Working
Answer
Keep the volumes and concentrations of both solutions, the flask, the cross and the depth of liquid the same. Bring both solutions to the intended starting temperature using a water bath. Record the actual mixture temperature and check how much it changes during each run; do not assume the water-bath temperature is the reaction temperature.
Read the data
Method
Turn the times into a rate trend.Reason
Each run reaches the same end-point, so a shorter time means a faster rate.
Working
Answer
The time falls from 120 s to 18 s as the temperature rises from 20 °C to 50 °C, roughly halving with each 10 °C rise. The rate increases.
Conclude and explain
Working
Answer
Between 20 °C and 50 °C, a higher temperature gives a faster rate. The particles have more kinetic energy on average, move faster and collide more often, and a greater fraction of collisions have enough energy to react. There are more successful collisions per unit time.
Quick check
Check your answer for the factor you change, a timed end-point, three things you keep the same, the falling times and both effects of temperature on collisions. The data cover 20–50 °C only, so they say nothing about other temperatures. Repeat each temperature to see how much the results vary.
Check a comparison that changed two variables
Spot the mistake 3
Two things changed at once
Learner working
A student adds 0.10 g of zinc granules to 50 cm³ of dilute sulfuric acid, then 0.20 g of zinc powder to another 50 cm³ of the same acid at the same temperature. The acid is in excess. The powder gives hydrogen faster and makes twice the volume. The student writes: (1) “The powder has a larger surface area.” (2) “The powder run was faster, so smaller particles caused the faster rate.” (3) “Smaller particles also make more product.” Find the first error and say how to fix the investigation.
Find the first unsupported step
Show solution step by step
Check the plan
Method
Compare the two runs before reading the results.
Reason
A conclusion about one factor needs every other factor kept the same.
Working
Answer
Particle size changed, and the mass of zinc doubled.
Find the first error
Method
Reject statement (2).Reason
Statement (1) is true. But twice the mass of zinc also exposes more surface, so the faster rate cannot be put down to particle size alone.
Working
Answer
Two variables changed, so the test cannot isolate particle size.
Explain the extra hydrogen
Method
Reject statement (3).Reason
Zinc is the reactant that runs out, and there is twice as much of it.
Working
Answer
With complete reaction in excess acid, twice the zinc gives twice the hydrogen. Particle size changes how quickly it forms.
Fix the plan
Working
Answer
Repeat with 0.10 g of zinc powder. With equal masses, compare the rates over the same interval. The final volumes should then match.
Common mistake
Check that your correction names both changed variables, identifies statement (2) as the first unsupported inference, explains the doubled gas amount and uses equal zinc masses. A faster result does not show which factor caused it. First check that only one factor changed; then read rate from the gradient and amount from the final volume.
The thiosulfate reaction produces sulfur dioxide, so follow the teacher’s small-scale, ventilated procedure and do not smell the gas. Keep the view of the cross, mixing method and timing start consistent. Record the reaction mixture’s temperature, not just the water-bath setting.
Check your understanding 2 Separate concentration from temperature
Two runs use equal masses of the same-sized marble chips with 50 cm³ of acid, which is in excess. Run 1 uses 1.0 mol/dm³ acid at 20 °C. Run 2 uses 2.0 mol/dm³ acid from a bottle that stood on a sunny windowsill, at 31 °C. Run 2 is faster. Is this a fair test of concentration? Explain how to improve it.
Compare both changed factors, how you would control temperature, the remaining controls and which reactant runs out.
Show answer
No: concentration and temperature both changed, and each affects the rate, so the faster rate cannot be put down to concentration alone. Bring both acids to the same temperature, for example in a water bath, and measure it. Keep marble mass and particle size, acid volume and temperature the same; change only concentration. Keeping the acid in excess means marble is the limiting reactant, so the final gas amounts can be compared.
Check your understanding 3 Use the mixture readings
A student warms only the thiosulfate in a 40 °C water bath, then adds acid from the bench at 22 °C. The mixture cools approximately steadily from 36 °C at the start to 35 °C at the end, and the cross disappears in 45 s. Estimate a representative mixture temperature for plotting against 45 s, to the nearest 0.1 °C. Choose where it comes from, then explain why 40 °C is wrong and how to improve the method in the next task.
Show answer
Plot about 35.5 °C, the mean of the mixture’s starting and final temperatures: the reaction happens in the mixture. The 40 °C setting was never the reaction temperature, because the cooler acid lowered the temperature on mixing. Warm both solutions in the bath before mixing, then measure the mixture’s temperature during each run.
Check your understanding 4 Fix the temperature method
In the run above, only thiosulfate was warmed to 40 °C before acid at 22 °C was added. Explain why the bath setting is unsuitable for the graph, how to improve the procedure, and when the two-reading average is a reasonable estimate.
Show answer
Compare fixed end-point times
For the same visible change, a shorter time means a faster average change. Use 1/t as a relative rate index, where t is the disappearance time in seconds. Halving the time doubles this index. It compares progress to a fixed end-point, rather than measuring an initial tangent gradient. Keep the flask, liquid depth, cross and viewing conditions the same so the end-point is comparable.
Check your understanding 5 Compare disappearance times
A cross disappears in 80 s in one run and 40 s in another. What must be the same before the times show which run is faster? For a fixed end-point, use the relative rate index 1/time to calculate the 40 s run’s index divided by the 80 s run’s index. Can the times alone identify the factor changed?
Show the steps Optional working
Show answer
The end-point must represent the same cloudiness: keep the cross, viewing conditions, flask and liquid depth the same. The 40 s run reaches that fixed change in half the time. Its relative rate index is twice as large: (1/40)/(1/80) = 2. This is an end-point comparison, not an initial tangent rate. Several factors could shorten the time, so the times alone do not identify the cause.
Check your understanding 6 Equal times, different liquid depths
Two students time the thiosulfate reaction using identical flasks and the same cross. Both record 30 s, but one mixture is 1 cm deep and the other is 2 cm deep. A student concludes that the rates are equal. Evaluate that conclusion and plan a fair comparison of concentration using this method. Include what you would change, what you would measure, how you would keep the end-point comparable and what you would repeat.
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Common mistakes
- Changing two factors at once, such as the mass and the size of a solid, and crediting the result to one.
- Naming a controlled variable without saying how it is kept the same.
- Recording the water-bath setting instead of the temperature of the reaction mixture.
- Comparing disappearance times when the liquid depths or viewing end-points differ.
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Syllabus and review details
- SEC G3 Combined Science Chemistry component 2027 · 2027
Chemistry content structure and subject content, PDF pages 26–36
Last reviewed: