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9476 / 2027
Reaction Kinetics

Lesson 2 of 5

Concentration-time graphs and half-life

Recognise constant loss versus a constant fraction lost.

A-Level 9476 (2026-2027)

  • I can explain the central kinetic terms.Syllabus

    Syllabus 9476, 8(a). Explain the central kinetic terms.

  • I can construct and use simple rate equations.Syllabus

    Syllabus 9476, 8(b). Construct and use simple rate equations.

  • I can understand and use first-order half-life.Syllabus

    Syllabus 9476, 8(c). Understand and use first-order half-life.

Zero order loses equal amounts; first order loses equal fractions

Only first-order half-life stays constant as concentration falls.

Zero-order concentration-time graph

Concentration falls in a straight line from 0.20 mol per cubic decimetre to zero over 60 seconds. The constant negative gradient means a constant disappearance rate.

Illustrative zero-order behaviour while the same rate law remains valid. Going from 0.20 to 0.10 takes 30 s; going from 0.10 to 0.05 takes only 15 s. Do not extend the line below zero concentration.

First-order concentration-time graph

Concentration falls from 0.20 to 0.10, 0.050 and 0.025 mol per cubic decimetre at 0, 20, 40 and 60 seconds. The curve becomes less steep, but each halving takes 20 seconds.

Illustrative first-order decay. The magnitude of the gradient decreases with concentration; the half-life stays 20 s under the same conditions.

The half-life is the time taken for a reactant concentration to fall to half its starting value. In a first-order reaction, each equal time interval removes the same fraction of the reactant. So the half-life stays the same, whatever the starting concentration. To show first order on a graph, find several halvings in a row and check that each takes the same time. One halving alone is not enough evidence.

Worked exampleUse repeated half-lives without an integrated rate equation

A first-order reactant has a half-life of 12.0 min. Starting at 0.160 mol dm-3, find its concentration after 36.0 min and the percentage reacted.

Try it first. Then show the worked stepsHide the worked steps
  1. 36.0/12.0 = 3 half-lives.
  2. Halve three times: 0.160 → 0.0800 → 0.0400 → 0.0200 mol dm-3.
  3. Fraction remaining = 1/8; fraction reacted = 7/8.
Answer

0.0200 mol dm-3 remains; 87.5% has reacted. The first-order model approaches zero but does not reach it after a finite number of half-lives.

Try it step by step

Work it out in steps: Number of half-lives, then Concentration remaining, then Percentage reacted.

  1. Problem 1. A first-order reactant has a half-life of 20.0 s. It starts at 0.400 mol dm-3. Find its concentration after 60.0 s and the percentage reacted, to 3 significant figures.
  2. Problem 2. A first-order reactant has a half-life of 15.0 min. It starts at 0.240 mol dm-3. Find the percentage reacted after 30.0 min, to 3 significant figures.
  3. Problem 3. A first-order reactant has a half-life of 8.00 min. It starts at 0.800 mol dm-3. Find the number of half-lives in 32.0 min, then the concentration and the percentage reacted to 3 significant figures.
Show the answer

Problem 1.

Number of half-lives: 60.0 / 20.0 = 3 half-lives.

Concentration remaining: 0.400 -> 0.200 -> 0.100 -> 0.0500 mol dm-3.

Percentage reacted: 1/8 remains, so 7/8 has reacted: 87.5%.

Problem 2.

Number of half-lives: 30.0 / 15.0 = 2 half-lives.

Concentration remaining: 0.240 -> 0.120 -> 0.0600 mol dm-3.

Percentage reacted: 1/4 remains, so 3/4 has reacted: 75.0%.

Problem 3.

Number of half-lives: 32.0 / 8.00 = 4 half-lives.

Concentration remaining: 0.800 -> 0.400 -> 0.200 -> 0.100 -> 0.0500 mol dm-3.

Percentage reacted: 1/16 remains, so 15/16 has reacted: 93.75%, which is 93.8% to 3 significant figures.

Sometimes one reactant is kept at an almost constant concentration, for example because it is in large excess. Then the decay you observe shows only how the rate depends on the other reactant. Say this condition before you give an overall order for the whole reaction. You do not need integrated rate equations for this syllabus.