(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 step by step, we will follow the principles of first-order reactions and the rate equation.

### Step 1: Identify the given data
- Initial concentration of A, \( [A]_0 = 0.1 \, M \)
- Final concentration of A, \( [A]_t = 0.025 \, M \)
- Time, \( t = 40 \, \text{minutes} \)

### Step 2: Use the first-order rate equation to find the rate constant \( k \)
The formula for the rate constant \( k \) for a first-order reaction is given by:

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

Substituting the values into the equation:

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

### Step 3: Calculate the logarithm
Calculate \( \frac{0.1}{0.025} \):

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

Now, calculate \( \log(4) \):

\[ \log(4) = 2 \log(2) \approx 2 \times 0.3010 = 0.6020 \]

### Step 4: Substitute back to find \( k \)
Now substitute \( \log(4) \) back into the equation for \( k \):

\[ k = \frac{2.303}{40} \times 0.6020 \]

Calculating this gives:

\[ k \approx \frac{2.303 \times 0.6020}{40} \approx 0.0347 \, \text{min}^{-1} \]

### Step 5: Find the rate of reaction when concentration of A is \( 0.001 \, M \)
The rate of a first-order reaction is given by:

\[ \text{Rate} = k \times [A] \]

Substituting the values:

\[ \text{Rate} = 0.0347 \, \text{min}^{-1} \times 0.001 \, M \]

Calculating this gives:

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

### Final Answer
The rate of reaction of A when the concentration of A is \( 0.001 \, M \) is:

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

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