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Course homeModule II-08 · Recommended before: Before kinetics problem sets
Kinetics, integrated rate laws and Arrhenius
Rate constants have units that depend on reaction order. First-order half-life depends on k, not initial concentration.
Learn the method
t₁/₂ = 0.693/k (first order); ln k = −Eₐ/(RT)+ln A
- Identify reaction order from the stated rate law or experimental evidence.
- Choose the appropriate integrated equation.
- Keep time units consistent with k.
- Use natural logarithms for first-order decay.
- For Arrhenius comparisons, use inverse absolute temperatures.
Common mistake: Do not use a first-order half-life equation for an unspecified zero- or second-order process.
Worked example
Problem: Calculate a first-order half-life when k = 0.0500 s⁻¹.
Step-by-step solution: t₁/₂ = ln 2 / k = 0.693147 / 0.0500 = 13.9 s.
Check the units and whether the answer is reasonable before continuing.
More applications of this mathematics
Try each extension before expanding its solution.
Rate-law order
If doubling [A] increases rate fourfold with all else unchanged, the reaction is second order in A (2² = 4).
Arrhenius temperature comparison
For Ea = 50.0 kJ/mol, T₁ = 300 K and T₂ = 310 K, ln(k₂/k₁) = −Ea/R(1/T₂−1/T₁) ≈ 0.646; k₂/k₁ ≈ 1.91.
Zero- and second-order plots
For a zero-order reaction, [A] vs t is linear with slope −k. For a second-order reaction in one reactant, 1/[A] vs t is linear with slope +k.
Independent practice — five questions
Suggested mastery threshold: four out of five. Incorrect answers display a worked explanation after submission.
Supporting reference
OpenStax Chemistry 2e / NIST — reference for this subject. This lesson’s wording, examples and questions are original.
Applied practice: Integrated rate laws and half-life
These additional, ungraded problems cover skills beyond the scored five-question check above. Try each before expanding its solution. Your existing score and progress records are unchanged.
Challenge 1. A first-order reaction decreases from 0.200 M to 0.0500 M with k = 0.0350 min⁻¹. Find the elapsed time.
Show worked solution for challenge 1
Use t = ln([A]₀/[A]ₜ) ÷ k = ln(4.00) ÷ 0.0350 min⁻¹ = 39.6 min.
Challenge 2. A first-order process has a half-life of 12.0 min. What fraction remains after 36.0 min?
Show worked solution for challenge 2
Three half-lives have elapsed. Remaining fraction = (1/2)³ = 1/8 = 12.5%.
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