Question

A possible reaction for the degradation of the pesticide ddt to a less harmful compound was simulated in the laboratory. the reaction was found to be first-order, with k = 4.0 x 10⁻⁸ s⁻¹ at 25°c. what is the half-life for the degradation of ddt in this experiment, in years?

Answer 1

To calculate the half-life of DDT degradation, which is a first-order reaction, use the formula t₁/₂ = 0.693 / k and the provided rate constant. After calculation and conversion from seconds to years, the half-life is approximately 0.55 years.

Step-by-step explanation:

The student's question concerns the calculation of the half-life of a first-order reaction specifically for the degradation of DDT to a less harmful compound. According to the given rate constant (k = 4.0 x 10⁻⁸ s⁻¹) at 25°C, we need to utilize the formula for the half-life of a first-order reaction, which is t₁/₂ = 0.693 / k. Substituting the rate constant into the formula yields t₁/₂ in seconds, which can then be converted into years for practical understanding.

By placing the given values into the equation:

t₁/₂ = 0.693 / (4.0 x 10⁻⁸ s⁻¹)

t₁/₂ ≈ 1.7325 x 10⁷ seconds

To convert this value into years, we use the conversion factor that there are 31,536,000 seconds in a year (60 seconds/minute × 60 minutes/hour × 24 hours/day × 365 days/year):

t₁/₂ ≈ (1.7325 x 10⁷ seconds) / (31,536,000 seconds/year)

t₁/₂ ≈ 0.55 years

Therefore, the half-life for the degradation of DDT in this experiment is approximately 0.55 years.