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 biochemical oxygen demand (BOD) exerted at time $t$ is calculated using the formula: \[Y_t = Y \left( 1 - 10^{-K \cdot t} \right)\] where: - $Y_t$ represents the BOD exerted at time $t$. - $Y$ is the ultimate BOD. - $K$ is the reaction rate constant (base 10). - $t$ is the time in days.

Step 2: Substitution of given values.
Given that after 4 days, the BOD exerted is 0.75 times the ultimate BOD: \[Y_t = 0.75 Y\] Substituting this into the formula from Step 1: \[0.75 = 1 - 10^{-4K}\] Rearranging the equation to solve for the exponential term: \[10^{-4K} = 0.25\]

Step 3: Solving for $K$.
To solve for $K$, we take the logarithm of both sides of the equation: \[-4K \log 10 = \log 0.25\] This simplifies to: \[-4K = \log_{10}(0.25)\] Isolating $K$: \[K = -\frac{\log_{10}(0.25)}{4}\] Calculating the numerical value: \[K = \frac{0.602}{4} \approx 0.171\]

Step 4: Conclusion.
Therefore, the reaction rate constant $K$ is approximately 0.171 per day. The correct option is (C).