The specific rate constant for a first order reaction is `1 times 10^–3 sec^–1`. If the initial concentration of the reactant is 0.1 mole/lt, the rate of reaction `mol l^–1 sec^–1` is:

Options

A. `10^-4`
B. `10^-3`
C. `10^-2`
D. `10^-1`

Answer

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

### Step-by-Step Solution:

1. **Identify the given values**:
- Specific rate constant (k) = \(1 \times 10^{-3} \, \text{sec}^{-1}\)
- Initial concentration of the reactant ([A]) = \(0.1 \, \text{mol/L}\)

2. **Write the rate law for a first-order reaction**:
The rate of a first-order reaction is given by the equation:
\[ \text{Rate} = k \times [A] \]
where:
- Rate is the rate of reaction in \( \text{mol/L/sec} \)
- k is the specific rate constant
- [A] is the concentration of the reactant

3. **Substitute the known values into the rate law**:
\[ \text{Rate} = (1 \times 10^{-3} \, \text{sec}^{-1}) \times (0.1 \, \text{mol/L}) \]

4. **Perform the multiplication**:
\[ \text{Rate} = 1 \times 10^{-3} \times 0.1 = 1 \times 10^{-4} \, \text{mol/L/sec} \]

5. **Final result**:
The rate of the reaction is:
\[ \text{Rate} = 1 \times 10^{-4} \, \text{mol/L/sec} \]

### Conclusion:
The rate of the reaction is \(1 \times 10^{-4} \, \text{mol/L/sec}\).

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