Explain first-order half-life independence
Show why a first-order reactant has the same half-life at different starting concentrations.
Section 26 · Reaction kinetics
A Level; Cambridge International Chemistry 9701, 2028–2030 syllabus, Topic 26. Rate laws are empirical unless an elementary step is explicitly stated. Written practice is self-checked
Before you begin
- Use AS collision theory, concentration, graph gradients, and balanced reaction equations.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What determines first-order half-life at fixed temperature?
What you will learn
- Relate first-order rate to reactant concentration
- Recognise half-life as time to halve a concentration
- Explain why first-order half-life is independent of initial concentration
A first-order rate falls in proportion to concentration
For a first-order reaction, rate = k[A]. As [A] decreases, the rate decreases in the same proportion. The half-life is the time for [A] to fall from a chosen value to half that value.
The first-order relation gives t₁/₂ = 0.693/k, which contains k but not the initial concentration. At a fixed temperature, k is constant, so repeated halvings take equal time.
Equal half-lives do not mean constant rate
The concentration halves in each successive half-life, and the rate also halves because rate = k[A]. The reaction continues more slowly; only the time interval for each halving stays constant for first-order kinetics.
This independence is characteristic of first order. Do not generalise it to zero- or second-order reactions.
Pause and explain
If a first-order starting concentration doubles, what happens to its half-life at fixed temperature?
Worked example
A first-order reaction has k = 0.020 s⁻¹. Compare the half-life starting from 0.80 and 0.20 mol dm⁻³.
Show the worked solution
- Use t₁/₂ = 0.693/k.
- Substitute the same k for both starts.
- Check that initial concentration is absent.
Answer Both starting concentrations have t₁/₂ = 0.693/0.020 = 34.7 s at this temperature.
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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 first-order reaction halves from 0.40 to 0.20 mol dm⁻³ in 50 s. How long should it take to halve from 0.20 to 0.10 mol dm⁻³ at the same temperature?
Give me a hint
Successive first-order half-lives are equal.
Compare my reasoning
- Identify the first half-life as 50 s.
- Keep temperature and k fixed.
- Apply the same interval to the next halving.
It should take 50 s, assuming the same first-order conditions and temperature.
Look for these in your work
- Halving interval identified
- First-order property applied
- Conditions stated
Explain why the reaction rate is lower after one first-order half-life even though the next half-life lasts the same time.
Give me a hint
Use rate = k[A] and the concentration change.
Compare my reasoning
- [A] is halved after one half-life.
- At fixed k, rate = k[A] is also halved.
- The subsequent time to halve is still 0.693/k.
The rate falls with [A], so after one half-life it is half as large. The next halving still takes the same time because first-order t₁/₂ depends only on k.
Look for these in your work
- Rate decrease explained
- Concentration dependence used
- Equal half-life distinguished
A learner says a reaction's half-life proves it is first order because the term half-life is used. Correct the reasoning.
Give me a hint
Half-life is defined for many reactions; its concentration dependence is the evidence.
Compare my reasoning
- Half-life means time to halve concentration.
- Measure it from different starting concentrations.
- Independence from starting concentration supports first-order behaviour.
The term half-life alone does not identify order. If the measured half-life is independent of starting concentration under fixed conditions, that supports first-order kinetics.
Look for these in your work
- Definition stated
- Comparison proposed
- Evidence interpreted
Common mistakes
- Saying the reaction rate stays constant
- Including initial concentration in the first-order half-life equation
- Applying equal half-lives to all orders
- Changing k while temperature is fixed
What can you explain now?
What determines first-order half-life at fixed temperature?
Compare with the explanation
The rate constant k
t₁/₂ = 0.693/k and has no initial concentration term.
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Make it stick
Why do successive first-order concentration halvings take equal time while the reaction rate falls?
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.