For consecutive first order reaction
`Aoverset(k_(1))(to)Boverset(k_(2))(to)C`, at 300 K
`k_(1)=2xx10^(-3)s^(-1)` and `k_(2)=5xx10^(-5)s^(-1)`
The time which [B] will be maximum is

Options

A. `189.2` s
B. `1892` s
C. 0 s
D. `oo`

Explanation

To determine the time at which the concentration of B is maximum in the consecutive first-order reaction \( A \overset{k_1}{\rightarrow} B \overset{k_2}{\rightarrow} C \), we can use the formula for the time at which the concentration of an intermediate is maximum in consecutive reactions.

### Step-by-Step Solution:

1. **Identify the Rate Constants**:
- Given:
- \( k_1 = 2 \times 10^{-3} \, s^{-1} \)
- \( k_2 = 5 \times 10^{-5} \, s^{-1} \)

2. **Use the Formula for Maximum Concentration of B**:
- The time at which the concentration of B is maximum (\( T_{max} \)) can be calculated using the formula:
\[ T_{max} = \frac{2.303}{k_2} \log_{10} \left( \frac{k_1}{k_2} \right) \]

3. **Calculate \( \frac{k_1}{k_2} \)**:
- First, calculate the ratio:
\[ \frac{k_1}{k_2} = \frac{2 \times 10^{-3}}{5 \times 10^{-5}} = \frac{2}{5} \times 10^{2} = 40 \]

4. **Calculate the Logarithm**:
- Now, calculate the logarithm:
\[ \log_{10}(40) \approx 1.602 \]

5. **Substitute Values into the Formula**:
- Now substitute \( k_2 \) and the logarithm into the formula:
\[ T_{max} = \frac{2.303}{5 \times 10^{-5}} \times 1.602 \]

6. **Calculate \( T_{max} \)**:
- Calculate \( T_{max} \):
\[ T_{max} = \frac{2.303 \times 1.602}{5 \times 10^{-5}} \approx \frac{3.688}{5 \times 10^{-5}} \approx 73760 \, seconds \]

7. **Final Calculation**:
- Simplifying gives:
\[ T_{max} \approx 73760 \, seconds \]

### Final Answer:
The time at which the concentration of B is maximum is approximately **73760 seconds**.