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. 276 s
B. 414 s
C. 552 s
D. 690 s

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 half-life**:
The half-life (t_half) of the substance is given as 138 seconds.

2. **Determine the initial and final concentrations**:
- Initial concentration (C_initial) = 1.28 mg L^(-1)
- Final concentration (C_final) = 0.04 mg L^(-1)

3. **Calculate the number of half-lives required**:
We can find how many times we need to halve the initial concentration to reach the final concentration.
- First half-life: 1.28 mg L^(-1) → 0.64 mg L^(-1)
- Second half-life: 0.64 mg L^(-1) → 0.32 mg L^(-1)
- Third half-life: 0.32 mg L^(-1) → 0.16 mg L^(-1)
- Fourth half-life: 0.16 mg L^(-1) → 0.08 mg L^(-1)
- Fifth half-life: 0.08 mg L^(-1) → 0.04 mg L^(-1)

From this, we can see that it takes 5 half-lives to go from 1.28 mg L^(-1) to 0.04 mg L^(-1).

4. **Calculate the total time required**:
The total time required (t_total) can be calculated using the formula:
\[ t_{total} = n \times t_{half} \]
where \( n \) is the number of half-lives.
\[ t_{total} = 5 \times 138 \text{ seconds} = 690 \text{ seconds} \]

5. **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**.