Rate constant k=2.303 `min^(-1)` for a particular reaction. The initial concentration of the reactant is 1 mol`//`litre then rate of reaction after 1 minute is:

Options

A. 2.303 M `min^(-1)`
B. 0.2303 M `min^(-1)`
C. 0.1 M `min^(-1)`
D. None of these

Answer

To solve the problem, we need to calculate the rate of a first-order reaction after 1 minute, given the rate constant and the initial concentration of the reactant.

### Step-by-Step Solution:

1. **Identify the Order of the Reaction**:
- The rate constant \( k \) is given as \( 2.303 \, \text{min}^{-1} \). The unit of \( k \) indicates that this is a first-order reaction because the unit is time inverse (min\(^{-1}\)).

2. **Write the Rate Law for a First-Order Reaction**:
- For a first-order reaction, the rate can be expressed as:
\[ \text{Rate} = k \cdot [A] \]
- Where \( [A] \) is the concentration of the reactant.

3. **Use the First-Order Kinetics Equation**:
- The concentration of the reactant at time \( t \) can be calculated using the equation:
\[ [A] = [A_0] e^{-kt} \]
- Here, \( [A_0] \) is the initial concentration, \( k \) is the rate constant, and \( t \) is the time.

4. **Substitute the Given Values**:
- The initial concentration \( [A_0] \) is \( 1 \, \text{mol/L} \).
- The rate constant \( k \) is \( 2.303 \, \text{min}^{-1} \).
- The time \( t \) is \( 1 \, \text{min} \).

5. **Calculate the Concentration After 1 Minute**:
- Substitute the values into the equation:
\[ [A] = 1 \cdot e^{-2.303 \cdot 1} \]
- Calculate \( e^{-2.303} \):
- Since \( e^{2.303} \approx 10 \), we have:
\[ [A] = 1 \cdot \frac{1}{10} = 0.1 \, \text{mol/L} \]

6. **Calculate the Rate of Reaction**:
- Now, substitute \( [A] \) back into the rate law:
\[ \text{Rate} = k \cdot [A] = 2.303 \cdot 0.1 \]
- Calculate the rate:
\[ \text{Rate} = 0.2303 \, \text{mol/L/min} \]

### Final Answer:
The rate of the reaction after 1 minute is \( 0.2303 \, \text{mol/L/min} \).