Understanding Half-Life Problems in Chemistry

Sep 18, 2024

Lecture Notes: Half-Life Problems in Chemistry

Introduction to Half-Life Problems

  • Focus on half-life problems encountered in chemistry courses.
  • Example: Iodine-131 with a half-life of 8 days.

Concept of Half-Life

  • Definition: The time it takes for half of a substance to decay.
  • Application: Understanding how much of a sample remains after a given period.

Solving Half-Life Problems: Methods

  1. Conceptual Understanding
    • Example: 200 grams of Iodine-131.
    • Calculate remaining sample after multiple half-lives.
      • 8 days: 100 grams remain.
      • 16 days: 50 grams remain.
      • 24 days: 25 grams remain.
      • 32 days: 12.5 grams remain.
  2. Using Equations
    • Calculate rate constant (K):
      • Formula: ( K = \frac{ln(2)}{\text{half-life}} )
      • For Iodine-131: K = 0.08664.
    • Final amount formula:
      • ( A_f = A_i \times e^{-Kt} )
    • For Iodine-131 (200g, 32 days): Result ≈ 12.5 grams.

Problem Example: Sodium-24

  • Given: Sodium-24 with a half-life of 15 hours, initial amount 800 grams.
  • Find: Time for 750 grams to decay.
    • Process: Calculate remaining Sodium-24 over half-lives.
    • Remaining amount after four half-lives: 50 grams.
    • Time calculation: 15 hours per half-life, total 60 hours.

Using Equation

  • Rate Constant K: ( K = \frac{ln(2)}{15} = 0.04621 )
  • Solve for Time T:
    • Equation: ( ln\left(\frac{50}{800}\right) = -KT )
    • Result: T ≈ 60 hours.

Problem Example: Oxygen-15

  • Given: Half-life of 2 minutes.
  • Find: Fraction remaining after 5 half-lives.
    • Calculation based on percentages and conversions to fractions.
    • Result: Fraction remaining = ( \frac{1}{32} ).

Problem Example: Decay of Element X

  • Given: 512 grams decay to 4 grams in 35 days.
  • Find: Half-life of Element X.
    • Calculation involves counting number of half-lives (7 in total).
    • Half-life Calculation: 5 days.

Using Equation

  • Rate Constant K: Use ( ln\left(\frac{4}{512}\right) = -KT )
  • Result: K ≈ 0.13863
  • Solve for Half-life: ( \text{Half-life} = \frac{ln(2)}{K} )
    • Result: 5 days.

Conclusion

  • Understanding both conceptual and equation-based approaches is important.
  • Equations allow handling of less straightforward numbers effectively.

Additional Resources

  • Encourage practice with problems and review of concepts.
  • Appreciate watching the video and have a great day.