For the first-order reaction `(C=C_(0)e^(-k_(1)^(t)))` and `T_(av)=k_(1)^(-1)`. After two average lives concentration of the reactant is reduced to :

Options

A. 0.25
B. 0.75
C. `(100)/(e )%`
D. `(100)/(e^(2))%`

Explanation

To solve the problem, we need to determine the concentration of the reactant after two average lives for a first-order reaction. The concentration of the reactant at time \( t \) is given by the equation:

\[ C_t = C_0 e^{-k_1 t} \]

Where:
- \( C_t \) is the concentration at time \( t \)
- \( C_0 \) is the initial concentration
- \( k_1 \) is the rate constant
- \( t \) is the time

### Step 1: Understand the average life
The average life \( T_{av} \) is given as \( T_{av} = k_1^{-1} \). This means that the average life is the reciprocal of the rate constant.

### Step 2: Calculate the time after two average lives
Since we are interested in the concentration after two average lives, we can express this time as:

\[ t = 2 T_{av} = 2 k_1^{-1} \]

### Step 3: Substitute the time into the concentration equation
Now, we substitute \( t \) into the concentration equation:

\[ C_t = C_0 e^{-k_1 (2 k_1^{-1})} \]

This simplifies to:

\[ C_t = C_0 e^{-2} \]

### Step 4: Assume an initial concentration
For simplicity, let's assume the initial concentration \( C_0 = 100 \) (this is arbitrary and can be any value, but it makes calculations easier).

### Step 5: Calculate the concentration after two average lives
Now, substituting \( C_0 \) into the equation:

\[ C_t = 100 e^{-2} \]

### Step 6: Calculate the numerical value
To find \( C_t \), we need to calculate \( e^{-2} \):

\[ C_t = 100 \times \frac{1}{e^2} \approx 100 \times 0.1353 \approx 13.53 \]

### Step 7: Calculate the percentage concentration remaining
To find the percentage of the initial concentration remaining, we can use the formula:

\[ \text{Percentage remaining} = \left( \frac{C_t}{C_0} \right) \times 100 \]

Substituting the values:

\[ \text{Percentage remaining} = \left( \frac{13.53}{100} \right) \times 100 \approx 13.53\% \]

### Final Answer
The concentration of the reactant after two average lives is approximately **13.53%** of the initial concentration.

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