Question

An underground hazardous waste storage tank is leaking. The contaminant concentration directly beneath the site is 0.5 mg/L. The contaminant is travelling at an effective rate of 0.4 m per day towards a water well which is 2 km away. Assume that the degradation of the contaminant follows a first order reaction, and the initial concentration of the contaminant becomes half in 10 years. In this case, the contaminant concentration expected at the well under steady state conditions is _______ mg/L (rounded off to two decimal places).

Hint:

For first order reactions, the degradation rate constant can be calculated from the half-life, and the concentration at a distance can be calculated based on time and travel velocity.

Answer 1

Correct Answer: 0.18

For a first order reaction, the concentration of the contaminant \( C(t) \) after time \( t \) is given by: \[ C(t) = C_0 e^{-kt}, \] where \( C_0 \) is the initial concentration, \( k \) is the degradation constant, and \( t \) is time. The contaminant concentration becomes half in 10 years, so: \[ \frac{C_0}{2} = C_0 e^{-k \cdot 10}, \] which simplifies to: \[ e^{-k \cdot 10} = \frac{1}{2}. \] Taking the natural logarithm of both sides: \[ -10k = \ln \left( \frac{1}{2} \right), \] \[ k = \frac{\ln 2}{10} = 0.0693 \, \text{year}^{-1}. \] Now, the contaminant is travelling at 0.4 m per day, and the well is 2000 m away. The time \( t \) for the contaminant to reach the well is: \[ t = \frac{2000 \, \text{m}}{0.4 \, \text{m/day}} = 5000 \, \text{days} = 13.7 \, \text{years}. \] The concentration of the contaminant at the well after 13.7 years is: \[ C(13.7) = 0.5 \, \text{mg/L} \cdot e^{-0.0693 \cdot 13.7} = 0.5 \, \text{mg/L} \cdot e^{-0.949} \approx 0.18 \, \text{mg/L}. \] Thus, the contaminant concentration expected at the well is \( 0.18 \, \text{mg/L} \).