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Concentration–Time Graphs and Half-life

Learn and apply Concentration–Time Graphs and Half-life in the published Chemistry course sequence.

  • GCE A-Level H2 Chemistry 9476-2027
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Concentration–Time Graphs and Half-life: Orientation

Concentration–time questions test whether you can recognise order and extract half-life from a graph. This lesson uses the graphical signature checks required by 9476 without integrated rate equations.

Link this page to Rate Equations, Orders, and Rate Constant and the Reaction Kinetics hub to keep mechanism and data interpretation consistent.

Definitions (Must Know)

A. Half-life, t_(1/2)

The half-life, t_(1/2), is the time taken for the concentration of a reactant to fall to half its value.

B. First-order reaction

A first-order reaction has rate proportional to concentration: rate = k[A]

C. Zero-order reaction

A zero-order reaction has rate independent of concentration: rate = k

D. Prescribed graphical boundary

Use concentration–time data to compare successive halving intervals. A constant half-life supports first-order behaviour. Integrated rate equations, logarithmic plots and formulae derived from them are explicitly not required by 9476.

Detailed Explanations

A. Recognising order from graph shape

Concentration–Time Graphs: Zero vs First Order (Example)

Concentration–Time Graphs: Zero vs First Order (Example). Zero order (linear), First order (curved) plotted as Concentration of A against Time, t.

Scroll across the graph to read all labels.

Concentration–Time Graphs: Zero vs First Order (Example). Zero order (linear), First order (curved) plotted as Concentration of A against Time, t.Concentration–Time Graphs: Zero vs First Order (Example). Zero order (linear), First order (curved) plotted as Concentration of A against Time, t.
Example shapes: zero order gives a straight-line decrease, while first order gives a curved (exponential) decrease.
Open full-size graph
View figure data
Values for Concentration–Time Graphs: Zero vs First Order (Example)
Time, t (s)Zero order (linear)First order (curved)
011
100.80.607
200.60.368
300.40.223
400.20.135
5000.082

B. Half-life method (from a concentration–time graph)

  1. Choose a concentration value and its half (e.g. 0.80 → 0.40).
  2. Read the times for those concentrations from the graph.
  3. The difference is one half-life.
  4. Repeat for another halving to check if the half-life stays the same (first-order signature).

C. Boundary check

Stop after identifying the order or reading a half-life when that is what the question asks. Do not calculate k from an integrated first-order relationship or transform the data into a logarithmic plot; those treatments are outside the explicit 9476 requirement.

Worked Examples

Modelled example 1

Identify first-order half-life evidence

Core

Problem

A reaction has the same half-life each time the reactant concentration halves. What order is supported, and what is the relevant graphical evidence?
Study the worked solution

  1. Name the measured feature

    Method

    Compare successive concentration-halving intervals.

    Reason

    Half-life is measured from one concentration to half that concentration, not always from time zero.

    Working

    Each successive halving takes the same time.

  2. Infer the order

    Method

    Identify constant half-life as a first-order signature.

    Reason

    For first-order behaviour, the same fraction is removed in equal time intervals.

    Working

    The evidence supports first order in the reactant.

Guided practice 2

Read a half-life interval

About 5 min

Problem

A concentration–time graph shows the reactant falling from 0.80 to 0.40 mol dm⁻³ between 10 s and 50 s. State the half-life represented by this interval.

Try this before viewing the solution

Hints

Hint 1: confirm the halving
0.40 is half of 0.80.
Hint 2: use a time difference
Subtract the time at the larger concentration from the time at the smaller concentration.
View solution step by step

  1. Locate the two times

    Method

    Use 10 s for 0.80 and 50 s for 0.40 mol dm⁻³.

    Reason

    These are the endpoints of one concentration halving.

    Working

    tₛₜₐᵣₜ = 10 s; t_end = 50 s.

  2. Calculate the interval

    Method

    Subtract the two graph times.

    Reason

    Half-life is an elapsed time.

    Working

    t_(1/2) = 50-10 = 40 s.

Common misconception 3

Correct a constant-rate claim

Find and correct the mistake

Learner claim

A learner says, “A first-order reaction has a constant half-life, so its reaction rate is constant.” Explain the error using successive halvings from 0.80 to 0.40 and then 0.40 to 0.20 mol dm⁻³.

Compare amount changes

In equal half-life intervals, the concentration decrease is

View solution step by step

  1. Compare the decreases

    Method

    Calculate each absolute concentration change.

    Reason

    Rate concerns amount or concentration change per unit time.

    Working

    First decrease: 0.40; second decrease: 0.20 mol dm⁻³.

  2. Correct the interpretation

    Method

    State that equal times remove equal fractions, not equal concentration amounts.

    Reason

    The smaller absolute decrease in the same time means the rate has fallen.

    Working

    Constant half-life is compatible with a decreasing first-order rate.

Examiner practice 4

Reach one quarter by successive half-lives

3 marks

Problem

A first-order concentration–time graph has a half-life of 35 s. How long does it take for concentration to fall from its initial value to one quarter? [3 marks]

Try this before viewing the solution

View solution step by step

  1. Count halvings

    1 mark

    Method

    Write the fractional sequence from 1 to 1/4.

    Reason

    Each first-order half-life halves the current concentration.

    Working

    1 → 1/2 → 1/4: two half-lives.

  2. Apply the half-life

    1 mark

    Method

    Multiply 35 s by two intervals.

    Reason

    The half-life is concentration-independent for first-order behaviour.

    Working

    t = 2(35 s).

  3. State the result

    1 mark

    Method

    Complete the multiplication.

    Reason

    The question asks for elapsed time from the initial value.

    Working

    t = 70 s.

Challenge 5

Interpret a straight concentration–time trace

Minimal support

Problem

A straight concentration–time line falls from 0.80 to 0.60 mol dm⁻³ in 20 s and then to 0.40 mol dm⁻³ after another 20 s. Identify the supported order and explain whether successive half-lives would be constant.

Try this before viewing the solution

Hints

Hint 1: compare equal-time changes
The same absolute decrease of 0.20 mol dm⁻³ occurs in each 20 s interval.
Hint 2: contrast fractional decreases
A later halving covers a smaller absolute concentration change than an earlier halving.
View solution step by step

  1. Identify the graph signature

    Method

    Recognise the constant negative gradient.

    Reason

    Equal concentration amounts are consumed in equal times.

    Working

    The data support zero-order behaviour.

  2. Predict half-life behaviour

    Method

    State that successive half-lives become shorter as concentration falls.

    Reason

    At constant gradient, halving a smaller concentration requires a smaller absolute decrease.

    Working

    Zero order has constant rate/gradient, not constant half-life.

Mind Stretchers

Mind stretcher 1Extension

A reactant’s concentration falls from 0.80 to 0.40 mol dm⁻³ in 30 s, and from 0.40 to 0.20 mol dm⁻³ in 30 s. What order is supported, and what further graphical check would strengthen the conclusion?

Show Hint

Measure successive intervals for the same fractional decrease, rather than judging only the curve shape.

Show Answer

Mark scheme:

  • The half-life is constant (30 s each halving), so first-order behaviour is supported.
  • Check another successive halving interval on the concentration–time graph and account for graph-reading uncertainty.

Mind stretcher 2: Testing a half-life claimExtension

Question. A concentration falls from 0.80 to 0.40 mol dm⁻³ in 24 s and from 0.40 to 0.20 mol dm⁻³ in 25 s. What order is supported, and why is the one-second difference not fatal?

Show Hint

Experimental data have uncertainty; ask whether the two halving intervals are effectively constant.

Show Answer

The nearly constant successive half-lives support first-order behaviour. A small difference can arise from measurement and graph-reading uncertainty; the evidence is approximate, not mathematically exact.