The concentration of a reactant X decreases from 0.1 M to 0.005 M in 40 min. If the reaction follows first order kinetics, the rate of the reaction when the concentration of X is 0.01 M will be

Options

A. `1.73 xx 10^(4) M min^(-1)`
B. `3.47 xx 10^(-4) M min^(-1)`
C. `3.47 xx 10^(-5) M min^(-1)`
D. `7.5 xx 10^(-4) M min^(-1)`

Explanation

To solve the problem step by step, we will follow the principles of first-order kinetics.

### Step 1: Identify the given data
- Initial concentration \( [X]_0 = 0.1 \, M \)
- Final concentration \( [X] = 0.005 \, M \)
- Time \( t = 40 \, \text{min} \)

### Step 2: Use the first-order rate constant formula
For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{2.303}{t} \log\left(\frac{[X]_0}{[X]}\right) \]
where:
- \( [X]_0 \) is the initial concentration,
- \( [X] \) is the final concentration,
- \( t \) is the time in minutes.

### Step 3: Substitute the values into the formula
Substituting the values into the equation:
\[ k = \frac{2.303}{40} \log\left(\frac{0.1}{0.005}\right) \]

### Step 4: Calculate the logarithm
First, calculate the ratio:
\[ \frac{0.1}{0.005} = 20 \]
Now calculate the logarithm:
\[ \log(20) = \log(2 \times 10) = \log(2) + \log(10) \]
Using known values:
\[ \log(2) \approx 0.3010 \quad \text{and} \quad \log(10) = 1 \]
Thus,
\[ \log(20) = 0.3010 + 1 = 1.3010 \]

### Step 5: Calculate \( k \)
Now substitute \( \log(20) \) back into the equation for \( k \):
\[ k = \frac{2.303}{40} \times 1.3010 \]
Calculating this gives:
\[ k = \frac{2.303 \times 1.3010}{40} \approx \frac{2.9963}{40} \approx 0.0749 \, \text{min}^{-1} \]

### Step 6: Calculate the rate of the reaction at \( [X] = 0.01 \, M \)
The rate of a first-order reaction is given by:
\[ \text{Rate} = k \times [X] \]
Substituting \( k \) and \( [X] \):
\[ \text{Rate} = 0.0749 \times 0.01 \]
Calculating this gives:
\[ \text{Rate} \approx 0.000749 \, M \, \text{min}^{-1} \quad \text{or} \quad 7.49 \times 10^{-4} \, M \, \text{min}^{-1} \]

### Final Answer
The rate of the reaction when the concentration of \( X \) is \( 0.01 \, M \) is approximately \( 7.49 \times 10^{-4} \, M \, \text{min}^{-1} \).

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