Half-Life Calculator

Calculate half-life and concentration for zero, first, and second-order reactions

Half-Life Calculator

Half-Life: The time required for the concentration of a reactant to decrease to half its initial value.
The formula varies with reaction order.

First Order: Rate = k[A]

t₁/₂ = ln(2)/k = 0.693/k | [A]ₜ = [A]₀e^(-kt)

s⁻¹ (or 1/time)

Key Differences:

  • • Zero Order: t₁/₂ depends on [A]₀ (decreases as reaction proceeds)
  • • First Order: t₁/₂ is constant (independent of concentration)
  • • Second Order: t₁/₂ depends on [A]₀ (increases as reaction proceeds)
  • • Radioactive decay is always first-order

Understanding Half-Life in Chemical Kinetics

Half-life (t₁/₂) is the time required for the concentration of a reactant to decrease to exactly one-half of its initial value. It's a fundamental concept in chemical kinetics and is particularly important in radioactive decay, pharmacokinetics, and environmental chemistry.

Half-Life Formulas by Reaction Order

Zero-Order Reactions

t₁/₂ = [A]₀ / (2k)
  • • Half-life depends on initial concentration
  • • t₁/₂ decreases as reaction proceeds
  • • Example: Photochemical reactions at high intensity

First-Order Reactions

t₁/₂ = ln(2) / k = 0.693 / k
  • • Half-life is constant (independent of concentration)
  • • Most common in nature (radioactive decay)
  • • Example: ¹⁴C decay, many drug eliminations

Second-Order Reactions

t₁/₂ = 1 / (k[A]₀)
  • • Half-life depends on initial concentration
  • • t₁/₂ increases as reaction proceeds
  • • Example: Dimerization reactions, some decompositions

Integrated Rate Laws

OrderRate LawIntegrated FormLinear Plot
Zero=k[A]ₜ=[A]₀-kt[A] vs t
FirstRate=k[A]ln[A]ₜ=ln[A]₀-ktln[A] vs t
SecondRate=k[A]²1/[A]ₜ=1/[A]₀+kt1/[A] vs t

Practical Example: First-Order Radioactive Decay

Problem: Carbon-14 has a half-life of 5,730 years. If a fossil contains 25% of the original ¹⁴C, how old is the fossil?

Given:

  • t₁/₂ = 5,730 years
  • [A]ₜ / [A]₀ = 0.25 (25% remaining)

Solution:

First, find k: k = 0.693 / t₁/₂

k = 0.693 / 5,730 = 1.209 × 10⁻⁴ yr⁻¹

Then, use: ln([A]ₜ/[A]₀) = -kt

ln(0.25) = -1.209 × 10⁻⁴ × t

-1.386 = -1.209 × 10⁻⁴ × t

t = 11,460 years

This is exactly 2 half-lives (5,730 × 2 = 11,460 years). After 2 half-lives, 25% remains (50% → 25%).

Common Half-Lives

Radioactive Isotopes

  • • ¹⁴C: 5,730 years
  • • ²³⁸U: 4.5 billion years
  • • ¹³¹I: 8 days (medical use)
  • • ²²²Rn: 3.8 days

Pharmaceuticals

  • • Aspirin: 2-3 hours
  • • Caffeine: 4-6 hours
  • • Ibuprofen: 2 hours
  • • Penicillin: 0.5-1 hour

Applications

  • ⚛️
    Nuclear Chemistry: Radioactive dating, nuclear medicine, waste management
  • 💊
    Pharmacology: Drug dosing schedules and elimination rates
  • 🌍
    Environmental Science: Pollutant degradation and atmospheric chemistry
  • 🧪
    Chemical Industry: Reaction optimization and shelf-life determination
  • 🏛️
    Archaeology: Carbon-14 dating of artifacts and fossils

⏱️Quick Reference

Zero Order:

t₁/₂ = [A]₀/(2k)

First Order:

t₁/₂ = 0.693/k

Second Order:

t₁/₂ = 1/(k[A]₀)

Key Property:

Only 1st order has constant t₁/₂

🔗Related Calculators

📐Related Formulas

🎯Where It's Used

  • ⚛️

    Nuclear Chemistry

    Radioactive decay dating

  • 💊

    Pharmacology

    Drug elimination rates

  • 🧪

    Chemical Kinetics

    Reaction rate studies

  • 🌍

    Environmental

    Pollutant degradation