Calculate reaction rates from experimental data
Calculate average rates from measurements and interpret a rate graph without overstating its precision.
AS Level; Cambridge International Chemistry 9701, 2025–2027 syllabus. Rate data and diagrams are illustrative unless a question supplies measurements. A diagram is qualitative unless labelled values are provided. Written practice is self-checked
Before you begin
- Rearrange simple ratios and use consistent units for time, volume, mass and concentration.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
How is average rate calculated from a measured quantity?
What you will learn
- Calculate average rate from experimental readings
- Use concentration, gas volume or mass data consistently
- Interpret graph gradients and units
Average rate uses a change divided by a time interval
For a measured quantity x, average rate over an interval is change in x divided by change in time. If product concentration rises from 0.10 to 0.34 mol dm⁻³ in 12 s, the average rate is (0.34 − 0.10)/12 = 0.020 mol dm⁻³ s⁻¹.
The measured variable sets the units: gas volume in cm³ over seconds gives cm³ s⁻¹; mass change in grams over minutes gives g min⁻¹; concentration in mol dm⁻³ over seconds gives mol dm⁻³ s⁻¹. State whether the quantity is increasing or a reactant is being consumed.
A graph gradient represents rate
On a graph of concentration or product amount against time, the gradient gives a rate. A secant between two data points gives an average rate over that interval. A tangent estimates an instantaneous rate at a selected time, if the graph and question require it.
Use two suitably separated points on a straight best-fit tangent for an initial or instantaneous gradient. A few joined experimental readings do not reveal an exact smooth curve or justify extra precision beyond the supplied measurements.
Pause and explain
What does the gradient of a product-volume–time graph represent?
Worked example
A reaction forms 36 cm³ of gas by 24 s. Find the average gas-production rate from 0 to 24 s.
Show the worked solution
- Change in gas volume = 36 − 0 = 36 cm³.
- Time interval = 24 − 0 = 24 s.
- Average rate = 36/24 cm³ s⁻¹.
Answer 1.5 cm³ s⁻¹ over the first 24 s.
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Test this topic →From guided practice to a new situation
Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.
A reactant concentration falls from 0.80 to 0.50 mol dm⁻³ in 15 s. Calculate its average disappearance rate.
Give me a hint
Use the magnitude of the decrease divided by the elapsed interval.
Compare my reasoning
- Concentration decreases by 0.80 − 0.50 = 0.30 mol dm⁻³.
- Divide 0.30 by 15 s.
- Report the rate with concentration-per-time units.
Average disappearance rate = 0.020 mol dm⁻³ s⁻¹.
Look for these in your work
- Concentration change correct
- Interval used
- Units included
Collected gas volumes are 9.0 cm³ at 5.0 s and 21.0 cm³ at 15.0 s. Find the average rate for that interval.
Give me a hint
Use differences between the two readings, not either reading alone.
Compare my reasoning
- Volume change = 21.0 − 9.0 = 12.0 cm³.
- Time change = 15.0 − 5.0 = 10.0 s.
- Rate = 12.0/10.0 cm³ s⁻¹.
1.20 cm³ s⁻¹ between 5.0 s and 15.0 s.
Look for these in your work
- Both changes calculated
- Correct interval used
- Rate and units reported
The supplied graph shows product volume increasing from 0 to 48 cm³ by 30 s, then remaining at 48 cm³ at 60 s. Compare the average rate in each 30 s interval and explain the later plateau.
Give me a hint
Calculate each interval separately and distinguish a plateau from zero product.
Compare my reasoning
- First interval: 48/30 = 1.6 cm³ s⁻¹.
- Second interval: (48−48)/30 = 0 cm³ s⁻¹ average gas production.
- The plateau means no additional gas was collected over that interval; it does not mean the vessel contains no product.
The first average rate is 1.6 cm³ s⁻¹; the second is 0 cm³ s⁻¹ because the measured volume did not increase.
Look for these in your work
- Intervals kept separate
- Both average rates calculated
- Plateau interpreted as no additional measured gas
Common mistakes
- Using final quantity divided by the wrong time interval
- Dropping the units
- Using a single later reading to claim the initial rate
- Mixing cm³ and dm³ or seconds and minutes
What can you explain now?
How is average rate calculated from a measured quantity?
Compare with the explanation
Change in quantity divided by change in time
Both the quantity change and the interval matter.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Which two differences do you calculate before reporting a measured rate and its units?
If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.