The rate constant of a first order reaction is `2.31 xx 10^(-2) s^(-1)`. What will be the time required for the initial concentration, 0.1M, of the reactant to be reduced to 0.05 M?

Explanation

To solve the problem, we will use the first-order reaction rate equation. The equation is given by:

\[ t = \frac{1}{k} \ln \left( \frac{[A_0]}{[A_t]} \right) \]

Where:
- \( t \) = time required
- \( k \) = rate constant
- \( [A_0] \) = initial concentration
- \( [A_t] \) = concentration at time \( t \)

### Step 1: Identify the given values
- Rate constant, \( k = 2.31 \times 10^{-2} \, s^{-1} \)
- Initial concentration, \( [A_0] = 0.1 \, M \)
- Final concentration, \( [A_t] = 0.05 \, M \)

### Step 2: Substitute the values into the equation
We need to find \( t \):

\[ t = \frac{1}{2.31 \times 10^{-2}} \ln \left( \frac{0.1}{0.05} \right) \]

### Step 3: Calculate the ratio of concentrations
Calculate \( \frac{[A_0]}{[A_t]} \):

\[ \frac{0.1}{0.05} = 2 \]

### Step 4: Calculate the natural logarithm
Now, find \( \ln(2) \):

\[ \ln(2) \approx 0.693 \]

### Step 5: Substitute the values into the equation
Substituting the values back into the equation:

\[ t = \frac{1}{2.31 \times 10^{-2}} \times 0.693 \]

### Step 6: Calculate the time
First, calculate \( \frac{0.693}{2.31 \times 10^{-2}} \):

\[ t \approx \frac{0.693}{0.0231} \approx 30 \, s \]

### Final Answer
The time required for the concentration to be reduced from 0.1 M to 0.05 M is approximately **30 seconds**.

---