Answer:
The reaction rate k is 0.0012563 (1/hour).
Step-by-step explanation:
We considered the reactions occurring in the plant as first order, and represented by this equation:
![y = L (1- e^(-kt))](https://img.qammunity.org/2020/formulas/chemistry/college/y0a2l6cr8wwl6u9m58i5n731oq3y7o80nz.png)
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
![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)](https://img.qammunity.org/2020/formulas/chemistry/college/volcbmoglroj8d4zcp451hixgwbgi2ir7t.png)
The reaction rate k is 0.0012563 1/hour.