The specific rate constant for a first order reaction is `60xx10^(-4) sec^(-1)`. If the initial concentration of the reactant is `0.01mol L^(-1)`, the rate is

Options

A. `60xx10^(-6)Msec^(-1)`
B. `36xx10^(-4)Msec^(-1)`
C. `60xx10^(-2)Msec^(-1)`
D. `36xx10^(-1)Msec^(-1)`

Answer

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

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Specific rate constant (k) = \(60 \times 10^{-4} \, \text{sec}^{-1}\)
- Initial concentration of the reactant \([A]_0\) = \(0.01 \, \text{mol L}^{-1}\)

2. **Write the Rate Equation for a First-Order Reaction**:
- The rate of a first-order reaction can be expressed as:
\[ \text{Rate} = k \times [A]_0 \]
where \(k\) is the specific rate constant and \([A]_0\) is the initial concentration of the reactant.

3. **Substitute the Given Values into the Rate Equation**:
- Substitute \(k\) and \([A]_0\) into the equation:
\[ \text{Rate} = (60 \times 10^{-4} \, \text{sec}^{-1}) \times (0.01 \, \text{mol L}^{-1}) \]

4. **Calculate the Rate**:
- Perform the multiplication:
\[ \text{Rate} = 60 \times 10^{-4} \times 0.01 \]
\[ \text{Rate} = 60 \times 10^{-6} \, \text{mol L}^{-1} \text{sec}^{-1} \]
\[ \text{Rate} = 6.0 \times 10^{-5} \, \text{mol L}^{-1} \text{sec}^{-1} \]

5. **Final Answer**:
- The rate of the reaction is \(6.0 \times 10^{-5} \, \text{mol L}^{-1} \text{sec}^{-1}\).