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 per litre the rate (in 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 find the rate of a first-order reaction given the specific rate constant (k) and the initial concentration of the reactant (A0).

### Step-by-Step Solution:

**Step 1: Identify the given values.**
- The specific rate constant (k) is given as \(1 \times 10^{-3} \, \text{sec}^{-1}\).
- The initial concentration of the reactant (A0) is given as \(0.1 \, \text{mol/L}\).

**Step 2: Write the rate equation for a first-order reaction.**
For a first-order reaction, the rate (R) can be expressed as:
\[ R = k \cdot [A_0] \]
where:
- \(R\) is the rate of the reaction,
- \(k\) is the specific rate constant,
- \([A_0]\) is the initial concentration of the reactant.

**Step 3: Substitute the known values into the rate equation.**
Substituting the values we have:
\[ R = (1 \times 10^{-3} \, \text{sec}^{-1}) \cdot (0.1 \, \text{mol/L}) \]

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

**Step 5: Write the final answer.**
Thus, the rate of the reaction is:
\[ R = 1 \times 10^{-4} \, \text{mol/L/sec} \]

### Conclusion:
The rate of the reaction is \(1 \times 10^{-4} \, \text{mol/L/sec}\), which corresponds to option 1 in the provided choices.

---