The rate constant of a first order reaction is `6.9xx10^(-3)s^(-1)`. How much time will it take to reduce the initial concentration to its `1//8^("th")` value?

Options

A. 100 s
B. 200 s
C. 300 s
D. 400 s

Explanation

To solve the problem of how much time it will take to reduce the initial concentration of a first-order reaction to its \( \frac{1}{8} \)th value, we can use the first-order reaction rate equation. Here are the steps to find the solution:

### Step-by-Step Solution:

1. **Understand the first-order reaction rate equation**:
The rate constant \( k \) for a first-order reaction is related to the concentration of the reactant at time \( t \) by the equation:
\[ k = \frac{1}{t} \log \left( \frac{C_0}{C_t} \right) \]
where:
- \( C_0 \) is the initial concentration,
- \( C_t \) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

2. **Set up the equation for the given problem**:
We want to find the time \( t \) when the concentration \( C_t \) is \( \frac{C_0}{8} \). Thus, we can rewrite the equation as:
\[ k = \frac{1}{t} \log \left( \frac{C_0}{\frac{C_0}{8}} \right) \]

3. **Simplify the logarithm**:
The expression inside the logarithm simplifies as follows:
\[ \frac{C_0}{\frac{C_0}{8}} = 8 \]
Therefore, we can rewrite the equation:
\[ k = \frac{1}{t} \log(8) \]

4. **Rearranging the equation to solve for \( t \)**:
Rearranging gives:
\[ t = \frac{\log(8)}{k} \]

5. **Substituting the values**:
We know that \( k = 6.9 \times 10^{-3} \, s^{-1} \) and \( \log(8) = \log(2^3) = 3 \log(2) \). Using \( \log(2) \approx 0.301 \):
\[ \log(8) = 3 \times 0.301 = 0.903 \]
Now substituting \( k \) and \( \log(8) \) into the equation:
\[ t = \frac{0.903}{6.9 \times 10^{-3}} \]

6. **Calculating \( t \)**:
\[ t \approx \frac{0.903}{0.0069} \approx 130.43 \, seconds \]

7. **Final Result**:
The time it will take to reduce the initial concentration to its \( \frac{1}{8} \)th value is approximately \( 130.43 \) seconds.