Concentration of the reactant in first order is reduced to `(1)/(e^(2))` after

Options

A. one natural life time
B. two natural life time
C. three natural life time
D. four natural life time

Answer

To solve the problem of determining the time taken for the concentration of a reactant in a first-order reaction to be reduced to \( \frac{1}{e^2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the first-order reaction**:
For a first-order reaction, the relationship between the concentration of the reactant at any time \( t \) and its initial concentration \( [A_0] \) is given by the equation:
\[ [A] = [A_0] e^{-kt} \]
where \( k \) is the rate constant.

2. **Setting up the equation**:
We know that the concentration is reduced to \( \frac{1}{e^2} \) after time \( t \). Therefore, we can write:
\[ \frac{[A_0]}{e^2} = [A_0] e^{-kt} \]

3. **Canceling the initial concentration**:
Since \( [A_0] \) appears on both sides of the equation, we can cancel it out (assuming \( [A_0] \neq 0 \)):
\[ \frac{1}{e^2} = e^{-kt} \]

4. **Taking the natural logarithm**:
To solve for \( kt \), we take the natural logarithm of both sides:
\[ \ln\left(\frac{1}{e^2}\right) = -kt \]
This simplifies to:
\[ -2 = -kt \]

5. **Solving for time \( t \)**:
Rearranging the equation gives us:
\[ kt = 2 \quad \Rightarrow \quad t = \frac{2}{k} \]

6. **Relating time to half-life**:
The half-life \( t_{1/2} \) for a first-order reaction is given by:
\[ t_{1/2} = \frac{0.693}{k} \]
We can express \( t \) in terms of half-lives:
\[ t = \frac{2}{k} = \frac{2 \times 0.693}{0.693} \cdot \frac{1}{k} = 2 \cdot t_{1/2} \]

### Final Answer:
The concentration of the reactant is reduced to \( \frac{1}{e^2} \) after \( 2 \) half-lives.