For a first order reaction , `A to` Products, the concentrations of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is:

Options

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

Explanation

To solve the problem, we need to find the rate of a first-order reaction when the concentration of A is 0.01 M, given that the concentration of A changes from 0.1 M to 0.025 M in 40 minutes.

### Step 1: Determine the rate constant (k)

For a first-order reaction, the integrated rate law is given by:

\[ \ln \left( \frac{[A]_0}{[A]} \right) = kt \]

Where:
- \([A]_0\) = initial concentration of A = 0.1 M
- \([A]\) = final concentration of A = 0.025 M
- \(t\) = time = 40 minutes = 2400 seconds (since we need to convert to seconds for standard units)
- \(k\) = rate constant

First, we need to calculate \(k\):

1. Substitute the values into the equation:

\[ \ln \left( \frac{0.1}{0.025} \right) = k \cdot 2400 \]

2. Calculate \(\frac{0.1}{0.025}\):

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

3. Now calculate \(\ln(4)\):

\[ \ln(4) \approx 1.386 \]

4. Substitute back to find \(k\):

\[ 1.386 = k \cdot 2400 \]

5. Solve for \(k\):

\[ k = \frac{1.386}{2400} \approx 0.0005775 \, \text{sec}^{-1} \]

### Step 2: Calculate the rate of reaction at [A] = 0.01 M

For a first-order reaction, the rate of reaction can be expressed as:

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

Where:
- \([A] = 0.01 \, \text{M}\)

Substituting the values:

\[ \text{Rate} = 0.0005775 \cdot 0.01 \]

Calculating the rate:

\[ \text{Rate} = 0.000005775 \, \text{M/sec} \, \text{or} \, 5.775 \times 10^{-6} \, \text{M/sec} \]

### Final Answer:
The rate of reaction when the concentration of A is 0.01 M is approximately \(5.775 \times 10^{-6} \, \text{M/sec}\).

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