The concentration of a solution is changed from 0.2 to 0.4, then what will be rate and rate constant. The reaction is of first order and rate constant is `K = 1xx10^(–6)` :

Options

A. `2xx10^-7,1xx10^-6`
B. `1xx10^-7,1xx10^6`
C. `4xx10^-7,1xx10^-6`
D. `2xx10^-3,1xx10^-3`

Answer

To solve the problem, we need to calculate the rate of a first-order reaction when the concentration of the reactant changes from 0.2 M to 0.4 M. We are also given the rate constant \( K = 1 \times 10^{-6} \, \text{s}^{-1} \).

### Step-by-Step Solution:

1. **Identify the formula for the rate of a first-order reaction**:
The rate of a first-order reaction can be expressed as:
\[ \text{Rate} = K \times [A] \]
where \( K \) is the rate constant and \( [A] \) is the concentration of the reactant.

2. **Substitute the given values**:
We know that the concentration of the reactant changes to 0.4 M. Therefore, we can substitute \( K \) and \( [A] \) into the formula:
\[ \text{Rate} = (1 \times 10^{-6} \, \text{s}^{-1}) \times (0.4 \, \text{M}) \]

3. **Calculate the rate**:
Now, we perform the multiplication:
\[ \text{Rate} = 1 \times 10^{-6} \times 0.4 = 4 \times 10^{-7} \, \text{mol/L/s} \]

4. **Identify the rate constant**:
The rate constant \( K \) is already provided as \( 1 \times 10^{-6} \, \text{s}^{-1} \).

5. **Final results**:
- Rate = \( 4 \times 10^{-7} \, \text{mol/L/s} \)
- Rate constant \( K = 1 \times 10^{-6} \, \text{s}^{-1} \)

### Conclusion:
The final answer is:
- Rate: \( 4 \times 10^{-7} \, \text{mol/L/s} \)
- Rate constant: \( 1 \times 10^{-6} \, \text{s}^{-1} \)