The half life of a substance in a certain enzyme catalyzed reaction is 138s. The time required for the concentration of the substance to fall from `1.28 mg L^(-1) to 0.04 mg L^(-1)` :

Options

A. 267s
B. 414s
C. 552s
D. 690s

Explanation

To solve the problem, we need to determine the time required for the concentration of a substance to decrease from 1.28 mg L^(-1) to 0.04 mg L^(-1) given that the half-life of the substance is 138 seconds.

### Step-by-Step Solution:

1. **Identify the Initial and Final Concentrations:**
- Initial concentration (C₀) = 1.28 mg L^(-1)
- Final concentration (C) = 0.04 mg L^(-1)

2. **Use the Half-Life Formula:**
The half-life (t₁/₂) is the time taken for the concentration of a substance to reduce to half of its initial value. The relationship between the initial concentration, final concentration, and half-life can be expressed using the formula:
\[ C = C₀ \left(\frac{1}{2}\right)^{n} \]
where \( n \) is the number of half-lives.

3. **Calculate the Number of Half-Lives (n):**
Rearranging the formula gives:
\[ n = \log_{1/2} \left(\frac{C₀}{C}\right) \]
Substituting the values:
\[ n = \log_{1/2} \left(\frac{1.28}{0.04}\right) \]

First, calculate the ratio:
\[ \frac{1.28}{0.04} = 32 \]
Now, calculate \( n \):
\[ n = \log_{1/2} (32) = \frac{\log_{10} (32)}{\log_{10} (0.5)} \]

We know that \( 32 = 2^5 \), so:
\[ \log_{10} (32) = 5 \cdot \log_{10} (2) \]
And since \( \log_{10} (0.5) = -\log_{10} (2) \):
\[ n = \frac{5 \cdot \log_{10} (2)}{-\log_{10} (2)} = -5 \]

Thus, \( n = 5 \).

4. **Calculate the Total Time (t):**
The total time required can be calculated using the formula:
\[ t = n \cdot t_{1/2} \]
Substituting the values:
\[ t = 5 \cdot 138 \text{ seconds} = 690 \text{ seconds} \]

### Final Answer:
The time required for the concentration of the substance to fall from 1.28 mg L^(-1) to 0.04 mg L^(-1) is **690 seconds**.

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