2) If the BOD; of a waste is 220 mg/l and the ultimate BOD is 320 mg/1, what is the rate constant?

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**Question 2: Calculating the Rate Constant**

The question involves determining the rate constant for a waste material sample. The details given are:

- The five-day Biochemical Oxygen Demand (BOD₅) of the waste is 220 mg/L.
- The ultimate Biochemical Oxygen Demand (BOD) is 320 mg/L.

The task is to calculate the rate constant for this decomposition process. 

**Explanation for Students:**

The BOD is a measure of the amount of oxygen required by microorganisms to decompose organic matter in water. The rate constant is an important parameter in water quality analysis as it helps predict how quickly organic pollutants will be broken down in a body of water. 

To calculate the rate constant (\(k\)), one can use the following formula: 

\[ L_t = L_0 (1 - e^{-kt}) \]

Where: 
- \(L_t\) is the BOD at time \(t\).
- \(L_0\) is the ultimate BOD.
- \(k\) is the rate constant.
- \(t\) is time, often in days for water treatment processes.

By rearranging the formula and using provided values, one can solve for the rate constant \(k\).
Transcribed Image Text:**Question 2: Calculating the Rate Constant** The question involves determining the rate constant for a waste material sample. The details given are: - The five-day Biochemical Oxygen Demand (BOD₅) of the waste is 220 mg/L. - The ultimate Biochemical Oxygen Demand (BOD) is 320 mg/L. The task is to calculate the rate constant for this decomposition process. **Explanation for Students:** The BOD is a measure of the amount of oxygen required by microorganisms to decompose organic matter in water. The rate constant is an important parameter in water quality analysis as it helps predict how quickly organic pollutants will be broken down in a body of water. To calculate the rate constant (\(k\)), one can use the following formula: \[ L_t = L_0 (1 - e^{-kt}) \] Where: - \(L_t\) is the BOD at time \(t\). - \(L_0\) is the ultimate BOD. - \(k\) is the rate constant. - \(t\) is time, often in days for water treatment processes. By rearranging the formula and using provided values, one can solve for the rate constant \(k\).
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