Measuring rates and reading reaction graphs
Choose a measurable change, record it against time and distinguish graph slope from its final level.
Section 6.2 · Rate of reaction
Exam scope and assessment notes
Core and Extended · Chemistry 0971, 2026–2028. Collision-theory explanations and method evaluation are taught in separate Extended lessons. Cambridge states that IGCSE 0971 has the same subject content as IGCSE 0620, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Read units and time intervals from a table or graph.
- Know that a gas can leave a reacting container or be collected.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
Which units describe gas volume gained per second?
What you will learn
- Describe gas-volume and mass-change methods for following suitable reactions.
- Calculate average rate and interpret initial slope and final plateau.
Follow a quantity that changes with the reaction
A gas-forming reaction can be followed by collecting gas in a suitable gas syringe and recording volume at known times. Alternatively, for an appropriate gas that escapes, record the reacting container's mass against time; the decrease corresponds to material leaving rather than atoms disappearing. The reaction, apparatus and teacher-supervised method must fit the measurement being used.
Start timing consistently with mixing and take readings at regular intervals. Average rate over an interval is the change in the chosen quantity divided by elapsed time: gas-volume increase in cm³/s, for example, or the magnitude of a mass decrease in g/s. This measured rate is tied to the chosen quantity and is not automatically a concentration rate in mol/dm³/s.
Slope tells speed; the plateau tells the observed total
On a collected-gas-volume versus time graph, a steeper slope means faster gas production over that part of the reaction. The curve usually becomes less steep as reactants are used and approaches a plateau when no further gas is being collected. Initial rate is inferred from the starting gradient, not the final volume divided by the entire time when a changing rate is being compared.
Two complete reactions can reach the same plateau at different times, giving equal observed final gas volumes with different rates. A mass-loss graph slopes down instead: a faster mass loss gives a steeper downward slope, and the positive rate magnitude uses how much mass was lost per unit time. Read labels before comparing graphs; an energy–reaction-progress diagram is not a gas-volume–time graph.
Pause and explain
Two complete gas-collection curves end at the same volume, but one gets there sooner. What is supported?
Worked example
A collected-gas table gives 0 cm³ at 0 s, 24 cm³ at 20 s and 36 cm³ at 40 s. Find the average gas-production rate in each interval and compare them.
Show the worked solution
- From 0 to 20 s, rate is (24−0)/(20−0)=1.2 cm³/s.
- From 20 to 40 s, rate is (36−24)/(40−20)=0.60 cm³/s.
- The later interval has half the earlier average rate, consistent with a decreasing graph slope.
Answer 1.2 cm³/s initially over the stated interval, then 0.60 cm³/s.
Try a topic test
Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.
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.
Curve B in the illustrative graph collects 15 cm³ gas by 10 s and 27 cm³ by 20 s. Find the average rate between those readings and name the measurements needed.
Give me a hint
Use the difference in gas volumes, not the total 27 cm³.
Compare my reasoning
- Gas gained is 27−15=12 cm³.
- Elapsed time is 20−10=10 s.
- Average interval rate is 12/10=1.2 cm³/s; a gas-volume measurement and timing are required.
1.2 cm³/s between 10 s and 20 s.
Look for these in your work
- Both differences use the same interval.
- The result has volume-per-time units.
In a mass-loss record, the container reads 100.00 g at 0 s, 99.60 g at 20 s and 99.40 g at 40 s. Calculate both interval rate magnitudes and describe the graph trend.
Give me a hint
Mass lost is a positive magnitude even though the graph slopes downward.
Compare my reasoning
- First interval loses 0.40 g in 20 s: 0.020 g/s.
- Second interval loses 0.20 g in 20 s: 0.010 g/s.
- The downward gradient becomes less steep; the later mass-loss rate is half the earlier one.
0.020 g/s then 0.010 g/s; the downward curve becomes less steep.
Look for these in your work
- Mass loss and graph direction are distinguished.
- The rate comparison follows the interval data.
Two equal-mass carbonate experiments with excess acid collect a final 48 cm³ gas. Curve A reaches its plateau at 30 s; curve B at 60 s. Compare their total-time average rates, then explain why these values are not exact initial rates.
Give me a hint
Use each total divided by its own time, while recognising that slope changes during reaction.
Compare my reasoning
- A's total-time average is 48/30=1.6 cm³/s.
- B's is 48/60=0.80 cm³/s, with the same collected total.
- Initial rates require the initial gradients; the full-time averages combine faster and slower parts of each reaction.
Full-time averages: A 1.6 cm³/s, B 0.80 cm³/s; initial rates require initial gradients.
Look for these in your work
- Total amount and average speed are compared separately.
- An average is not presented as an initial slope.
Common mistakes
- Using a final plateau as the reaction's speed.
- Ignoring which quantity and units are on the vertical axis.
- Calling a negative mass gradient a negative amount of reaction.
- Taking the full-time average as an exact initial rate.
What can you explain now?
A reacting container's mass falls from 102.40 g to 102.10 g in 30 s. Find the average mass-loss rate magnitude.
Compare with the explanation
0.010 g/s
Mass lost is 0.30 g; its positive rate magnitude is 0.30/30=0.010 g/s.
After trying it yourself, choose your next review. This is your self-assessment.
Your review choice appears on Today. Sign in to sync it across devices.
Make it stick
Describe one suitable gas-volume and one mass-change method. Draw their expected graph directions, label units and calculate a rate from two new readings.
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.