Question

A stream leaving a sewage pond (containing 80 mg/L of sewage) moves as a plug with a velocity of 40 m/hr. A concentration of 50 mg/L is measured 5,000 m downstream. What is the 1st order decay rate constant in the stream?

Answer 1

Answer:Decay rate constant,k = 0.00376/hr

Explanation:

IsT Order Rate of reaction is given as

In At/ Ao = -Kt

where [A]t is the final concentration at time t and [A]o is the inital concentration at time 0, and k is the first-order rate constant.

Initial concentration = 80 mg/L

Final concentration = 50 mg/L

Velocity = 40 m/hr

Distance= 5000 m

Time taken = Distance / Time

5000m / 40m/hr = 125 hr

In At/ Ao = -Kt

In 50/80 = -Kt

-0.47 = -kt

- K= -0.47 / 125

k = 0.00376

Decay rate constant,k = 0.00376/hr