75% of a first order reaction was completed in 32 minutes. When was 50% of the reaction completed ?

Options

A. 24 minutes
B. 16 minutes
C. 8 minutes
D. 4 minutes

Explanation

To solve the problem, we need to determine the time taken for 50% of a first-order reaction to be completed, given that 75% of the reaction was completed in 32 minutes.

### Step-by-Step Solution:

1. **Understanding the Reaction Completion**:
- If 75% of the reaction is completed, it means that 25% of the reactant remains.
- Therefore, if the initial concentration of the reactant is \( A_0 \), the concentration remaining after 75% completion is:
\[ A_t = A_0 - 0.75 A_0 = 0.25 A_0 \]

2. **Using the First-Order Reaction Formula**:
- For a first-order reaction, the rate constant \( k \) can be calculated using the formula:
\[ k = \frac{1}{t} \ln\left(\frac{A_0}{A_t}\right) \]
- Here, \( t = 32 \) minutes and \( A_t = 0.25 A_0 \).

3. **Substituting Values to Find \( k \)**:
- Substitute \( A_t \) into the formula:
\[ k = \frac{1}{32} \ln\left(\frac{A_0}{0.25 A_0}\right) = \frac{1}{32} \ln\left(\frac{1}{0.25}\right) = \frac{1}{32} \ln(4) \]

4. **Finding the Time for 50% Completion**:
- For 50% completion, \( A_t = 0.5 A_0 \).
- Now, we can use the same formula to find the time \( t_{1/2} \) for 50% completion:
\[ k = \frac{1}{t_{1/2}} \ln\left(\frac{A_0}{0.5 A_0}\right) = \frac{1}{t_{1/2}} \ln(2) \]

5. **Equating the Two Expressions for \( k \)**:
- From the previous calculations, we have:
\[ \frac{1}{32} \ln(4) = \frac{1}{t_{1/2}} \ln(2) \]

6. **Solving for \( t_{1/2} \)**:
- Rearranging gives:
\[ t_{1/2} = \frac{32 \ln(2)}{\ln(4)} \]
- Since \( \ln(4) = 2 \ln(2) \), we can simplify:
\[ t_{1/2} = \frac{32 \ln(2)}{2 \ln(2)} = 16 \text{ minutes} \]

### Final Answer:
The time taken for 50% of the reaction to be completed is **16 minutes**.