Question

A sample of waste-water has 4 day 20$^\circ$C B.O.D. value of 75% of the final B.O.D. The rate constant K (to the base 10) per day will be:

• A. 0.151
• B. 0.161
• C. 0.171
• D. 0.181

Hint:

For BOD kinetics, if 75% of BOD is exerted in 4 days, the rate constant can be quickly approximated using $K \approx 0.17$ (base 10).

Answer 1

The Correct Option is C

Step 1: Formula for BOD at time $t$.
The BOD exerted at time $t$ is given by: \[ Y_t = Y \left( 1 - 10^{-K \cdot t} \right) \] where, - $Y_t$ = BOD exerted at time $t$, - $Y$ = ultimate BOD, - $K$ = reaction rate constant (base 10), - $t$ = time in days.

Step 2: Substitution of given values.
It is given that after 4 days: \[ Y_t = 0.75 Y \] So, \[ 0.75 = 1 - 10^{-4K} \] \[ 10^{-4K} = 0.25 \]

Step 3: Solving for $K$.
\[ -4K \log 10 = \log 0.25 \] \[ -4K = \log_{10}(0.25) \] \[ K = -\frac{\log_{10}(0.25)}{4} \] \[ K = \frac{0.602}{4} \approx 0.171 \]

Step 4: Conclusion.
Hence, the rate constant $K$ is 0.171 per day. Correct answer is (C).