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 m i n^(-1)`
B. `0.692//24 m i n^(-1)`
C. `2.303 //24 log ( 1//8) mi n^(=1)`
D. `2.303 // 24 log ( 8 //1) mi n^(-1)`

Answer

To find the rate constant of the decomposition process that follows first-order kinetics, we can use the integrated rate law for first-order reactions. Here’s a step-by-step solution:

### Step 1: Understand the First-Order Kinetics
For a first-order reaction, the integrated rate law is given by the equation:
\[ \ln \left( \frac{[A_0]}{[A_t]} \right) = k \cdot t \]
where:
- \([A_0]\) = initial concentration
- \([A_t]\) = concentration at time \(t\)
- \(k\) = rate constant
- \(t\) = time

### Step 2: Identify Given Values
From the problem, we know:
- The concentration is reduced to \(\frac{1}{8}\) of its initial value, so:
\[ [A_t] = \frac{[A_0]}{8} \]
- The time \(t\) is given as 24 minutes.

### Step 3: Substitute Values into the Equation
Substituting the known values into the integrated rate law:
\[ \ln \left( \frac{[A_0]}{\frac{[A_0]}{8}} \right) = k \cdot 24 \]
This simplifies to:
\[ \ln(8) = k \cdot 24 \]

### Step 4: Calculate \(\ln(8)\)
We can express 8 as \(2^3\):
\[ \ln(8) = \ln(2^3) = 3 \ln(2) \]

### Step 5: Substitute \(\ln(8)\) into the Equation
Now substitute \(\ln(8)\) back into the equation:
\[ 3 \ln(2) = k \cdot 24 \]

### Step 6: Solve for \(k\)
Rearranging the equation to solve for \(k\):
\[ k = \frac{3 \ln(2)}{24} \]
\[ k = \frac{\ln(2)}{8} \]

### Step 7: Calculate the Value of \(k\)
Using the approximate value of \(\ln(2) \approx 0.693\):
\[ k = \frac{0.693}{8} \approx 0.086625 \]

### Step 8: Final Answer
Thus, the rate constant \(k\) is approximately:
\[ k \approx 0.0866 \, \text{min}^{-1} \]