(A) follows first order reaction, (A) changes from 0.1 M to 0.025 M in 40 min. Find the rate of reaction of A when concentration of A is 0.001 M.

Options

A. `3.47 xx 10^(-4) M "min"^(-1)`
B. `3.47 xx 10^(-5) M "min"^(-1)`
C. `1.73 xx 10^(-4) M "min"^(-1)`
D. `1.73 xx 10^(-5) M "min"^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the type of reaction and the formula for the rate constant (k)
Since the reaction follows first-order kinetics, we can use the formula for the rate constant (k) for a first-order reaction:

\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]} \right) \]

where:
- \([A]_0\) is the initial concentration,
- \([A]\) is the final concentration,
- \(t\) is the time in minutes.

### Step 2: Substitute the known values into the formula
From the problem, we have:
- \([A]_0 = 0.1 \, M\)
- \([A] = 0.025 \, M\)
- \(t = 40 \, \text{minutes}\)

Substituting these values into the formula gives:

\[ k = \frac{2.303}{40} \log \left( \frac{0.1}{0.025} \right) \]

### Step 3: Calculate the logarithm
First, calculate the ratio:

\[ \frac{0.1}{0.025} = 4 \]

Now calculate the logarithm:

\[ \log(4) \approx 0.602 \]

### Step 4: Calculate the rate constant (k)
Now substitute this back into the equation for k:

\[ k = \frac{2.303}{40} \times 0.602 \]

Calculating this gives:

\[ k \approx \frac{2.303 \times 0.602}{40} \approx 0.0347 \, \text{min}^{-1} \]

### Step 5: Calculate the rate of reaction when \([A] = 0.001 \, M\)
The rate of reaction (R) for a first-order reaction is given by:

\[ R = k \cdot [A] \]

Substituting the values we have:

\[ R = 0.0347 \, \text{min}^{-1} \cdot 0.001 \, M \]

### Step 6: Calculate the final rate
Now calculate the rate:

\[ R = 0.0347 \times 0.001 = 0.0000347 \, M \, \text{min}^{-1} = 3.47 \times 10^{-5} \, M \, \text{min}^{-1} \]

### Conclusion
The rate of reaction of A when the concentration of A is 0.001 M is:

\[ R = 3.47 \times 10^{-5} \, M \, \text{min}^{-1} \]