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, we will use the first-order reaction rate equation. The rate constant (k) is given, and we need to find the time (t) it takes for the concentration to reduce to \( \frac{1}{8} \) of its initial value.

### Step-by-Step Solution:

1. **Identify the given values:**
- Rate constant, \( k = 6.9 \times 10^{-3} \, s^{-1} \)
- Initial concentration, \( A = A_0 \) (let's denote it as \( A_0 \))
- Final concentration, \( A_t = \frac{A_0}{8} \)

2. **Use the first-order kinetics formula:**
The formula for a first-order reaction is:
\[ t = \frac{1}{k} \ln \left( \frac{A_0}{A_t} \right) \]

3. **Substitute the values into the equation:**
Since \( A_t = \frac{A_0}{8} \), we can write:
\[ t = \frac{1}{k} \ln \left( \frac{A_0}{\frac{A_0}{8}} \right) \]
Simplifying the fraction:
\[ t = \frac{1}{k} \ln (8) \]

4. **Calculate \( \ln(8) \):**
We know that:
\[ \ln(8) = \ln(2^3) = 3 \ln(2) \approx 3 \times 0.693 = 2.079 \]

5. **Substitute \( \ln(8) \) back into the equation:**
\[ t = \frac{1}{6.9 \times 10^{-3}} \times 2.079 \]

6. **Calculate \( t \):**
\[ t = \frac{2.079}{6.9 \times 10^{-3}} \approx 301.4 \, seconds \]

7. **Round the answer:**
The time required to reduce the initial concentration to its \( \frac{1}{8} \) value is approximately \( 301 \, seconds \).

### Final Answer:
The time required is approximately **301 seconds**.

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