For a given reaction `A rarr ` Product, rate is `1xx10^(-4)M s^(-1)` when `[A]=0.01M` and rate is `1.41xx10^(-4)M s^(-1)` when `[A]=0.02 M`. Hence, rate law is `:`

Options

A. `-(d[A])/(dt)=k[A]^(2)`
B. `-(d[A])/(dt)=k[A]`
C. `-(d[A])/(dt)=(k)/(4)[A]`
D. `-(d[A])/(dt)=k[A]^(1//2)`

Explanation

To solve the problem, we need to derive the rate law for the reaction \( A \rightarrow \text{Product} \) using the given rates and concentrations of reactant A.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Rate \( R_1 = 1 \times 10^{-4} \, \text{M s}^{-1} \) when \( [A_1] = 0.01 \, \text{M} \)
- Rate \( R_2 = 1.41 \times 10^{-4} \, \text{M s}^{-1} \) when \( [A_2] = 0.02 \, \text{M} \)

2. **Write the Rate Law:**
For a reaction of the form \( A \rightarrow \text{Product} \), the rate law can be expressed as:
\[ R = k[A]^n \]
where \( k \) is the rate constant and \( n \) is the order of the reaction.

3. **Set Up the Equations:**
From the first set of data:
\[ R_1 = k[A_1]^n \implies 1 \times 10^{-4} = k(0.01)^n \quad \text{(1)} \]
From the second set of data:
\[ R_2 = k[A_2]^n \implies 1.41 \times 10^{-4} = k(0.02)^n \quad \text{(2)} \]

4. **Divide the Two Equations:**
To eliminate \( k \), divide equation (1) by equation (2):
\[ \frac{1 \times 10^{-4}}{1.41 \times 10^{-4}} = \frac{k(0.01)^n}{k(0.02)^n} \]
This simplifies to:
\[ \frac{1}{1.41} = \left(\frac{0.01}{0.02}\right)^n \]
\[ \frac{1}{1.41} = \left(\frac{1}{2}\right)^n \]

5. **Solve for \( n \):**
Taking logarithms on both sides:
\[ \log\left(\frac{1}{1.41}\right) = n \log\left(\frac{1}{2}\right) \]
Rearranging gives:
\[ n = \frac{\log\left(\frac{1}{1.41}\right)}{\log\left(\frac{1}{2}\right)} \]
Calculating the values:
\[ n \approx \frac{-0.1461}{-0.3010} \approx 0.485 \approx \frac{1}{2} \]

6. **Write the Final Rate Law:**
Now that we have \( n \), we can write the rate law as:
\[ R = k[A]^{1/2} \]
or in terms of the rate of change of concentration:
\[ -\frac{d[A]}{dt} = k[A]^{1/2} \]

### Final Answer:
The rate law for the reaction is:
\[ -\frac{d[A]}{dt} = k[A]^{1/2} \]