Find out time period of Ist order reaction. When reaction complete `(2)/(3)` rd. if the value of rate constant is `4.3xx10^(-4)`

Options

A. `0.0025xx10^(3)`sec
B. `0.25xx10^(3)sec`
C. `0.025xx10^(3)sec`
D. `2.5xx10^(3)sec`

Explanation

To solve the problem of finding the time period for a first-order reaction when two-thirds of the reaction is complete, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Reaction Completion**:
- We know that two-thirds of the reaction is complete. If we denote the initial concentration of the reactant as \( A \), then the amount of reactant remaining after the reaction is \( A - X \), where \( X \) is the amount reacted.
- Since two-thirds of the reaction is complete, we have:
\[ X = \frac{2}{3}A \]
Therefore, the concentration remaining is:
\[ A - X = A - \frac{2}{3}A = \frac{1}{3}A \]

2. **Using the First-Order Reaction Equation**:
- The integrated rate law for a first-order reaction is given by:
\[ k t = 2.303 \log \left( \frac{A}{A - X} \right) \]
- Substituting \( A - X = \frac{1}{3}A \) into the equation, we get:
\[ k t = 2.303 \log \left( \frac{A}{\frac{1}{3}A} \right) = 2.303 \log(3) \]

3. **Substituting the Values**:
- We know the rate constant \( k = 4.3 \times 10^{-4} \, \text{s}^{-1} \).
- Rearranging the equation for time \( t \):
\[ t = \frac{2.303 \log(3)}{k} \]
- Now substituting the value of \( k \):
\[ t = \frac{2.303 \log(3)}{4.3 \times 10^{-4}} \]

4. **Calculating the Logarithm**:
- The logarithm \( \log(3) \) is approximately \( 0.477 \).
- Thus, we can calculate:
\[ t = \frac{2.303 \times 0.477}{4.3 \times 10^{-4}} \]

5. **Final Calculation**:
- Performing the multiplication:
\[ t \approx \frac{1.098351}{4.3 \times 10^{-4}} \approx 2550.8 \, \text{s} \]
- Rounding this gives us:
\[ t \approx 2.55 \times 10^{3} \, \text{s} \]

### Conclusion:
The time period for the first-order reaction when two-thirds of the reaction is complete is approximately \( 2.55 \times 10^{3} \, \text{s} \).