A reaction that is of the first order with respect to reactant A has a rate constant `0.6"min"^(-1)`. If we state with [A] = 0.5 `mol.l^(-1)`, when would.A reach the value `0.05 mol.l^(-1)` ?

Options

A. 0.384 min
B. 0.15 min
C. 3 min
D. 3.84 min

Answer

To solve the problem, we will use the first-order reaction kinetics formula. The formula for the time required for a first-order reaction to reach a certain concentration is given by:

\[ t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]} \right) \]

Where:
- \( t \) is the time required,
- \( k \) is the rate constant,
- \( [A]_0 \) is the initial concentration,
- \( [A] \) is the final concentration.

### Step-by-Step Solution:

1. **Identify the given values:**
- Rate constant, \( k = 0.6 \, \text{min}^{-1} \)
- Initial concentration, \( [A]_0 = 0.5 \, \text{mol.l}^{-1} \)
- Final concentration, \( [A] = 0.05 \, \text{mol.l}^{-1} \)

2. **Substitute the values into the formula:**
\[ t = \frac{2.303}{0.6} \log \left( \frac{0.5}{0.05} \right) \]

3. **Calculate the logarithm:**
- First, calculate the ratio:
\[ \frac{0.5}{0.05} = 10 \]
- Now, find the logarithm:
\[ \log(10) = 1 \]

4. **Substitute the logarithm back into the equation:**
\[ t = \frac{2.303}{0.6} \cdot 1 \]

5. **Calculate the time:**
\[ t = \frac{2.303}{0.6} \approx 3.838 \, \text{min} \]

### Final Answer:
The time required for the concentration of A to decrease from \( 0.5 \, \text{mol.l}^{-1} \) to \( 0.05 \, \text{mol.l}^{-1} \) is approximately **3.84 minutes**.