The decomposition of a substance follows first order kinetics. If its concentration is reduced to 1/8 th of its initial value in 12 minutes, the rate constant of the decomposition system is

Options

A. `((2.303)/(12) log (1)/(8)) "min"^(-1)`
B. `((2.303)/(12) log 8) "min"^(-1)`
C. `((0.693)/(12)) "min"^(-1)`
D. `((1)/(12) log 8) "min"^(-1)`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the problem
We need to find the rate constant \( k \) for a first-order reaction where the concentration of the reactant decreases to \( \frac{1}{8} \) of its initial value in 12 minutes.

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

### Step 3: Define the variables
Let:
- The initial concentration \( A = Y \)
- The concentration after 12 minutes \( A - X = \frac{Y}{8} \)

This means that \( X = Y - \frac{Y}{8} = Y \left(1 - \frac{1}{8}\right) = Y \left(\frac{7}{8}\right) \).

### Step 4: Substitute the values into the formula
Now we can substitute these values into the formula for \( k \):
\[ k = \frac{2.303}{12} \log \left( \frac{Y}{\frac{Y}{8}} \right) \]

### Step 5: Simplify the logarithm
The expression inside the logarithm simplifies as follows:
\[ \frac{Y}{\frac{Y}{8}} = 8 \]
So, we have:
\[ k = \frac{2.303}{12} \log(8) \]

### Step 6: Calculate the value of \( k \)
Now we can express \( k \) as:
\[ k = \frac{2.303}{12} \cdot \log(8) \]
The logarithm of 8 can be calculated using the change of base formula or known logarithmic values:
\[ \log(8) = \log(2^3) = 3 \log(2) \]
Thus, we can write:
\[ k = \frac{2.303}{12} \cdot 3 \log(2) \]

### Step 7: Final expression for \( k \)
The final expression for the rate constant \( k \) is:
\[ k = \frac{2.303 \cdot 3 \log(2)}{12} \text{ min}^{-1} \]