Answer:
The reaction rate k is 0.0012563 (1/hour).
Explanation:
We considered the reactions occurring in the plant as first order, and represented by this equation:
[tex]y = L (1- e^{-kt})[/tex]
where y is the BOD at time t, L is the initial value of BOD and k is the reaction rate.
If we replaced with the values
y = 2 mg O2/l (1% of the initial value)
L = 200 mg 02/l
t = 8 hr
We can calculate k
[tex]y=L(1-e^{-kt})\\\\k=-(1/t)*ln(1-y/L) = -(1/8)*ln(1-2/200)=-(1/8)*(-0.01)=0.0012563 \, hour^{-1}[/tex]
The reaction rate k is 0.0012563 1/hour.