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:
Show Hint
- 0.151
- 0.161
- 0.171
- 0.181
The Correct Option is C
Solution and Explanation
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).
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