If the rate constant for a first order reaction is k, the time (t) required for the completion of 99% of the reaction is given by :

Options

A. `t=2.303//k`
B. `t=0.693//k`
C. `t=6.909//k`
D. `t=4.606//k`

Explanation

To find the time required for the completion of 99% of a first-order reaction, we can use the first-order reaction kinetics formula:

### Step-by-Step Solution:

1. **Identify the formula for first-order kinetics**:
The formula for the time \( t \) required for a first-order reaction to reach a certain percentage of completion is given by:
\[ t = \frac{2.303}{k} \log\left(\frac{[A_0]}{[A]}\right) \]
where:
- \( k \) is the rate constant,
- \( [A_0] \) is the initial concentration,
- \( [A] \) is the concentration at time \( t \).

2. **Determine the initial and final concentrations**:
Since we want to find the time for 99% completion of the reaction, we can set:
- \( [A_0] = 100 \) (initial concentration),
- \( [A] = 1 \) (since 99% of the reactant has reacted, only 1% remains).

3. **Substitute the values into the formula**:
Plugging the values into the formula gives us:
\[ t = \frac{2.303}{k} \log\left(\frac{100}{1}\right) \]

4. **Simplify the logarithm**:
The expression simplifies to:
\[ t = \frac{2.303}{k} \log(100) \]
Since \( 100 = 10^2 \), we can write:
\[ \log(100) = \log(10^2) = 2 \log(10) \]
Knowing that \( \log(10) = 1 \), we have:
\[ \log(100) = 2 \]

5. **Final calculation**:
Now substituting back into the equation:
\[ t = \frac{2.303}{k} \cdot 2 = \frac{4.606}{k} \]

### Conclusion:
The time required for the completion of 99% of the reaction is:
\[ t = \frac{4.606}{k} \]