The decomposition of a substance follows first order kinetics. If its conc. Is reduced to 1/8th of its initial value, in 24 minutes, the rate constant of decomposition process is

Options

A. `1//24 "min"^(-1)`
B. `0.692//24 "min"^(-1)`
C. `2.303//24 log (1//8) "min"^(-1)`
D. `(2.303)/(24) log(8//1) "min"^(-1)`

Explanation

To find the rate constant \( k \) for the first-order decomposition reaction, we can use the first-order rate equation:

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

Where:
- \([A]_0\) is the initial concentration,
- \([A]\) is the concentration at time \( t \),
- \( k \) is the rate constant,
- \( t \) is the time.

### Step 1: Identify the initial and final concentrations
Given that the concentration is reduced to \( \frac{1}{8} \) of its initial value, we can express this as:
- \([A]_0 = A\) (initial concentration)
- \([A] = \frac{A}{8}\) (final concentration after 24 minutes)

### Step 2: Substitute into the first-order rate equation
Substituting the values into the equation:

\[ \ln \left( \frac{A}{\frac{A}{8}} \right) = kt \]

This simplifies to:

\[ \ln(8) = kt \]

### Step 3: Calculate \( \ln(8) \)
We know that:

\[ \ln(8) = \ln(2^3) = 3 \ln(2) \]

Using the approximate value \( \ln(2) \approx 0.693 \):

\[ \ln(8) \approx 3 \times 0.693 = 2.079 \]

### Step 4: Substitute time into the equation
We know that \( t = 24 \) minutes. We need to convert this into seconds for consistency in units:

\[ t = 24 \text{ minutes} = 24 \times 60 = 1440 \text{ seconds} \]

### Step 5: Solve for \( k \)
Now we can substitute \( t \) into the equation:

\[ 2.079 = k \times 1440 \]

To find \( k \):

\[ k = \frac{2.079}{1440} \]

### Step 6: Calculate \( k \)
Now, performing the division:

\[ k \approx 0.00145 \text{ s}^{-1} \]

### Final Answer
The rate constant \( k \) for the decomposition process is approximately:

\[ k \approx 0.00145 \text{ s}^{-1} \]

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