Use first-order half-life in calculations
Calculate a first-order half-life or concentration after repeated half-lives.
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
If k is in min⁻¹, what unit does 0.693/k have?
What you will learn
- Use t₁/₂ = 0.693/k
- Calculate the remaining concentration after whole or fractional half-lives
- Relate elapsed time to a number of half-life intervals
Connect k to the half-life
For first-order kinetics, t₁/₂ = 0.693/k when k is in reciprocal time units. If k is s⁻¹, the half-life is in seconds; if k is min⁻¹, it is in minutes.
The same k must apply over the interval, so temperature and reaction conditions are assumed unchanged.
Track repeated halvings
After n half-lives, [A] = [A]₀(1/2)ⁿ. For an elapsed time t, the number of half-life intervals is t/t₁/₂; fractional intervals can also be handled with the first-order exponential relation if supplied or derived.
For a simple question, count the half-lives before calculating and check that the remaining concentration decreases rather than increases.
Pause and explain
After three first-order half-lives, what fraction of [A]₀ remains?
Worked example
For a first-order reaction with k = 0.0350 min⁻¹, calculate t₁/₂ and [A] after 60.0 min if [A]₀ = 0.80 mol dm⁻³.
Show the worked solution
- Calculate t₁/₂ = 0.693/0.0350.
- Find the number of half-lives in 60.0 min.
- Apply the repeated-halving relationship.
Answer t₁/₂ = 19.8 min; 60.0 min is about 3.03 half-lives, so [A] ≈ 0.80(1/2)³·⁰³ ≈ 0.098 mol dm⁻³.
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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.
Calculate the half-life for a first-order reaction with k = 0.0100 s⁻¹.
Give me a hint
Use 0.693/k and retain reciprocal-time units.
Compare my reasoning
- t₁/₂ = 0.693/0.0100.
- Calculate the quotient.
- State seconds.
t₁/₂ = 69.3 s.
Look for these in your work
- Equation selected
- Division correct
- Time unit stated
A first-order reactant begins at 1.20 mol dm⁻³. Find the amount remaining after four half-lives.
Give me a hint
Multiply by one-half four times.
Compare my reasoning
- Remaining fraction = (1/2)⁴.
- Calculate 1.20×1/16.
- State mol dm⁻³.
[A] = 0.0750 mol dm⁻³.
Look for these in your work
- Half-life count correct
- Fraction calculated
- Concentration unit stated
A first-order reaction has half-life 12 min. Estimate the time for [A] to decrease to one-eighth of its initial value.
Give me a hint
One-eighth is (1/2)³.
Compare my reasoning
- Identify three half-life intervals.
- Multiply 3 by 12 min.
- State the assumption of constant conditions.
Three half-lives are needed, so the time is 36 min at unchanged conditions.
Look for these in your work
- Fraction linked to halvings
- Elapsed time calculated
- Conditions held fixed
Common mistakes
- Using a half-life unit inconsistent with k
- Forgetting to divide elapsed time by half-life
- Doubling concentration after elapsed time
- Treating a fractional half-life as an exact integer
What can you explain now?
If k is in min⁻¹, what unit does 0.693/k have?
Compare with the explanation
Minutes
The reciprocal of min⁻¹ is minutes.
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Make it stick
How do k, half-life count and starting concentration combine in a first-order calculation?
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.