Chemical Kinetics · Lesson 6 of 10
Half-Life, Radioactive Decay and Pseudo-First-Order Behaviour
“Connect exponential decay to repeated halving intervals and explain why an excess reactant can hide a higher-order dependence.”
• Derive and apply the first-order half-life relationship. • Compare zero- and first-order halving intervals. • Calculate remaining fractions, dating estimates and specified-conversion times. • Derive an observed first-order coefficient under excess-reactant conditions and state its limits.
A half-life is a time for a specified fractional change
The time for the remaining amount or concentration to fall to one half of a stated starting value under the specified kinetic law and conditions.
For first-order decay, every period of the same length removes the same fraction of what is then present. Substituting [R]t=[R]₀/2 into the integrated equation cancels the initial concentration. This cancellation is the mathematical reason the first-order half-life is independent of starting concentration; it is not an additional assumption.
Problem
Find the half-life when k₁=5.5×10⁻¹⁴ s⁻¹.
- 1.t₁/₂=ln2/(5.5×10⁻¹⁴)=1.26027e+13 s≈1.26×10¹³ s.
- 2.Using 365.25 days per year gives approximately 3.99×10⁵ years, illustrating how a favourable change can still be very slow.
- 3.The small inverse-time constant represents a small fractional change per second; the large half-life follows from the reciprocal relation.
Problem
Compare the time for 99.9% consumption with one first-order half-life.
- 1.99.9% consumption leaves fraction f=0.001. Therefore t=ln1000/k₁.
- 2.Divide by t₁/₂=ln2/k₁. The rate constant cancels.
- 3.The ratio is ln1000/ln2=9.96578≈10, not exactly ten. At exactly ten half-lives, the fraction remaining is 1/1024≈0.000976563, giving 99.9023% consumed.
- 4.The difference is small but clarifies that rounding a convenient rule must not turn it into an exact identity.
| Feature | Zero-order disappearance | First-order disappearance |
|---|---|---|
| Concentration law | [R]t=[R]₀−k₀t | [R]t=[R]₀e⁻ᵏ¹ᵗ |
| First half-life | [R]₀/(2k₀) | ln2/k₁ |
| Dependence on initial concentration | Directly proportional | Independent at fixed k₁ |
| Successive halving intervals | Become shorter in the same run | Stay equal |
| Constant units | Concentration/time | Time⁻¹ |
| Ideal full-depletion extrapolation | Finite linear intercept; regime may change earlier | No exact zero at finite time |
Radioactive decay is an amount-loss process
Radioactive decay of a particular unstable nuclide under fixed nuclear conditions is described by a first-order statistical law in its number of undecayed nuclei. Each nucleus has a decay probability per time; the expected number decaying per time is proportional to how many remain. The same exponential mathematics applies to amount or mass when it tracks that parent nuclide. This does not make nuclear decay an ordinary chemical reaction.
For example, ²²⁶₈₈Ra → ⁴₂He + ²²²₈₆Rn conserves nucleon number 226=4+222 and atomic number 88=2+86. A half-life describes a population’s statistical decay, not a prediction of exactly when a particular nucleus will transform. Environmental chemistry may move a nuclide between locations without changing how much has decayed, so dating and retention calculations require additional assumptions.
Problem
A closed sample initially had the same carbon-14 reference abundance as its living counterpart. It now has 80% remaining. Use half-life 5730 years to estimate its decay age.
- 1.For the stipulated closed-sample model, f=0.80 and k₁=ln2/5730 years⁻¹. Therefore t=5730×ln(1/0.80)/ln2=1844.65 years.
- 2.The estimate is about 1.84×10³ years. It is not 20% of one half-life because exponential loss is nonlinear.
- 3.Actual dating also requires suitable sample history and calibration; the calculation isolates the simple decay model rather than establishing an exact historical date.
Problem
An initial 1.00 mg sample of ⁹⁰Sr has half-life 28.1 years. Calculate the parent amount after 10 and 60 years if there is no transfer or metabolic loss.
- 1.Use m(t)=m₀2^(−t/28.1). At 10 years, m=0.781397 mg.
- 2.At 60 years, m=0.227631 mg. The later amount is smaller because more than two half-lives have elapsed.
- 3.These are amounts of the undecayed parent under the stipulated retention assumption, not total mass of parent plus decay products or a model of real biological clearance.
An excess reactant can create an observed first-order law
Suppose a measured law is v=k[A][B]. If B is in such large excess that its fractional concentration change is negligible while A reacts, [B] can be treated as approximately constant. Combine k[B] into one observed coefficient k_obs. The time dependence then appears first order in A, even though varying B between separate runs changes the observed coefficient.
Acid-catalysed ethyl acetate hydrolysis, CH₃COOC₂H₅+H₂O⇌CH₃COOH+C₂H₅OH, can exhibit an observed first-order dependence on ester in a suitable water-rich medium at fixed acidity. Water activity and catalyst conditions are approximately fixed and absorbed into k_obs. The simple two-concentration illustration is not a universal mechanistic law for ester hydrolysis in every solvent or pH range. Acid is written above the reaction arrow because it is regenerated rather than consumed overall.
Problem
An illustrative hydrolysis starts with 0.010 mol ester and 10.0 mol water. Compare the maximum stoichiometric water change with its starting amount.
- 1.If 0.010 mol ester reacts by the 1:1 hydrolysis balance, 0.010 mol water is used and 9.990 mol remains.
- 2.The maximum fractional water change is 0.010/10.0=0.001, or 0.10%, small enough to motivate treating water amount as nearly constant if volume and medium effects are also stable.
- 3.The balance is an upper-bound consumption illustration. Reversible ester hydrolysis does not guarantee exactly complete conversion.
- 4.The acid-catalysed hydrolysis of sucrose, C₁₂H₂₂O₁₁+H₂O→C₆H₁₂O₆(glucose)+C₆H₁₂O₆(fructose), offers another pseudo-first-order regime in excess water at fixed acidity.
A first-order time trace in A while B is fixed cannot establish that the full reaction is zero order in B. Compare multiple B concentrations if the aim is to determine that dependence. Exhaustion of the excess reactant, changing acidity or a changing solvent can invalidate k_obs constancy.
Quick Quiz
Quiz
A first-order half-life depends on:
Explanation: t₁/₂=ln2/k₁; the initial amount cancels.
After four half-lives, the first-order remaining fraction is:
Explanation: Each half-life multiplies the remaining amount by one half, giving (1/2)⁴=1/16.
How does the time for 99% consumption compare with that for 90% consumption in a first-order process?
Explanation: The times are ln100/k and ln10/k; ln100=2ln10.
For v=k[A][B], B nearly constant in large excess gives:
Explanation: The approximately constant B factor is absorbed into the observed coefficient multiplying [A].
Does a first-order decay curve in ester alone prove no dependence on water or acidity?
Explanation: Pseudo-first-order kinetics can conceal concentration or activity factors held effectively constant during a run.
Practice Problems
- Calculate half-lives for first-order constants 200 s⁻¹,2.0 min⁻¹ and 4.0 year⁻¹.
- SO₂Cl₂ has first-order half-life 60 min. Calculate k in min⁻¹ and s⁻¹.
- A first-order reaction has k₁=60 s⁻¹. How long until only 1/16 remains?
- A first-order reactant is 30% consumed in 40 min. Find its half-life.
- Sucrose has observed first-order half-life 3.00 h in fixed acid conditions. What fraction remains after 8.00 h?
- Prove t₉₉=2t₉₀ and calculate the change in k_obs if an excess reactant concentration is doubled in v=k[A][B].
Apply ln2/k separately while preserving the time unit. The half-lives are 0.00346574 s, 0.346574 min and 0.173287 year. Conversion to a common unit is optional but must be explicit.
Key Takeaways
• First-order half-life is ln2/k₁ and is independent of the starting amount. • Repeated first-order halving intervals are equal, while zero-order ones shorten during a run. • For a remaining fraction f, time is ln(1/f)/k₁. • 99.9% consumption takes approximately, rather than exactly, ten half-lives. • Radioactive parent amounts follow exponential population decay under stated assumptions. • An effectively constant excess reactant can be absorbed into k_obs. • Pseudo-first-order behaviour does not prove a unimolecular mechanism or no dependence on the fixed species. • Hydrolysis calculations require attention to acidity, excess solvent and reversibility.