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K324 / 2027
Rate of Reactions

Lesson 2 of 5

Choose a changing quantity

Match the measurement to the reaction.

O-Level 6092 (2026) / SEC G3 K324 (2027)

  • I can interpret rate data.Syllabus

    Syllabus K324, 10(f). Interpret rate data

Choose a changing quantity

Match the measurement to the reaction.

Ways to follow progress
MeasurementSuitable situationRate evidence
Gas volume against timeGas-producing reaction; gas can be collectedVolume increase per unit time
Mass against timeA gas escapes from an open reaction vesselMass loss per unit time
Time to a fixed visible endpointA precipitate obscures a mark or a colour reaches a chosen endpointShorter time means a faster average approach to that same endpoint

For a gas syringe, check airtight connections and free plunger movement. For mass loss, prevent splashes so that lost liquid is not mistaken for escaped gas. Start timing consistently. A subjective colour or visibility endpoint is less precise than a clear instrumental measurement, so use the same criterion and repeat.

Worked exampleCalculate an average rate

A reaction produces 36 cm3 of gas in its first 40 s.

Try it first. Then show the worked stepsHide the worked steps
  1. Average rate = change in gas volume / time interval.
  2. 36/40 = 0.90 cm3 s-1.
  3. This is an average over 40 s; the rate may have fallen throughout that interval.
Answer

0.90 cm3 s-1. State both interval and units.

Try it step by step

Work it out in steps: Change in the measured quantity, then Time interval, then Average rate.

  1. Problem 1. A gas syringe reads 12.0 cm3 at 20 s and 42.0 cm3 at 80 s. Find the average rate between 20 s and 80 s, in cm3 s-1 to 3 significant figures.
  2. Problem 2. A flask on a balance loses carbon dioxide. Its mass is 120.84 g at 0 s and 120.24 g at 120 s. Find the average rate of mass loss, in g s-1 to 2 significant figures (the mass lost, 0.60 g, has only 2).
  3. Problem 3. A gas syringe reads 15.0 cm3 at 10 s and 39.0 cm3 at 50 s. Find the change in volume, the time interval and the average rate. Give the rate in cm3 s-1 to 3 significant figures.
Show the answer

Problem 1.

Change in the measured quantity: Change in volume = 42.0 - 12.0 = 30.0 cm3.

Time interval: Time interval = 80 - 20 = 60 s.

Average rate: Average rate = 30.0 / 60 = 0.500 cm3 s-1 over that interval.

Problem 2.

Change in the measured quantity: Mass lost = 120.84 - 120.24 = 0.60 g.

Time interval: Time interval = 120 - 0 = 120 s.

Average rate: Average rate = 0.60 / 120 = 0.0050 g s-1. Two significant figures, because 0.60 g has two.

Problem 3.

Change in the measured quantity: Change in volume = 39.0 - 15.0 = 24.0 cm3.

Time interval: Time interval = 50 - 10 = 40 s.

Average rate: Average rate = 24.0 / 40 = 0.600 cm3 s-1.

Check your understanding

Two experiments reach the same end-point, one in 20 s and one in 40 s. Which is faster?

  1. Neither, because both reached the very same end-point.
  2. The 40 s one, because it ran for longer and so reacted more.
  3. The 20 s one, because the same change happened in less time.
Show the answer

The 20 s one, because the same change happened in less time.

For the same amount of change, rate is proportional to 1/time. 1/20 is twice 1/40, so the 20 s experiment is twice as fast.