The specfic reaction rate constant for a first order reaction is `1xx10^(-3) s^(-1)` .If the initial concentration of the reactant is `0.1 ` mole per litre , the rate 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 using the given rate constant and initial concentration of the reactant.

### Step-by-Step Solution:

**Step 1: Identify the formula for the rate of a first-order reaction.**
The rate of a first-order reaction is given by the equation:
\[ R = k \cdot [A] \]
where:
- \( R \) is the rate of the reaction,
- \( k \) is the specific reaction rate constant,
- \([A]\) is the concentration of the reactant.

**Step 2: Substitute the given values into the formula.**
From the problem, we have:
- \( k = 1 \times 10^{-3} \, \text{s}^{-1} \)
- \([A] = 0.1 \, \text{mol/L}\)

Now we can substitute these values into the rate equation:
\[ R = (1 \times 10^{-3} \, \text{s}^{-1}) \cdot (0.1 \, \text{mol/L}) \]

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

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