If 75% of first order reaction is completed in 32 min, then 50% of the reaction would be complete in:

Options

A. 16min
B. 24min
C. 10min
D. 40min

Answer

To solve the problem, we need to find out how long it takes for 50% of a first-order reaction to complete, given that 75% of the reaction is completed in 32 minutes.

### Step-by-Step Solution:

1. **Understanding the Reaction Order**:
The reaction is first-order. For a first-order reaction, the time taken to reach a certain percentage completion can be calculated using the formula:
\[ T_x = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A]} \right) \]
where \( T_x \) is the time taken to reach x% completion, \( [A_0] \) is the initial concentration, and \( [A] \) is the concentration at time \( T_x \).

2. **Setting Up for 75% Completion**:
If 75% of the reaction is complete, then 25% remains. Therefore, we can express this as:
\[ [A] = [A_0] \times 0.25 \]
Plugging this into the formula gives:
\[ T_{75} = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A_0] \times 0.25} \right) = \frac{2.303}{k} \log(4) = \frac{2.303}{k} \times 2 \log(2) \]

3. **Setting Up for 50% Completion**:
For 50% completion, 50% of the reactant remains:
\[ [A] = [A_0] \times 0.5 \]
Thus, we can express this as:
\[ T_{50} = \frac{2.303}{k} \log \left( \frac{[A_0]}{[A_0] \times 0.5} \right) = \frac{2.303}{k} \log(2) \]

4. **Finding the Relationship Between T75 and T50**:
Now, we can set up the ratio of \( T_{50} \) to \( T_{75} \):
\[ \frac{T_{50}}{T_{75}} = \frac{\frac{2.303}{k} \log(2)}{\frac{2.303}{k} \log(4)} \]
Since \( \log(4) = 2 \log(2) \), we can simplify:
\[ \frac{T_{50}}{T_{75}} = \frac{\log(2)}{2 \log(2)} = \frac{1}{2} \]
Therefore, we find that:
\[ T_{50} = \frac{T_{75}}{2} \]

5. **Calculating T50**:
Given that \( T_{75} = 32 \) minutes, we can now calculate \( T_{50} \):
\[ T_{50} = \frac{32 \text{ min}}{2} = 16 \text{ min} \]

### Final Answer:
Thus, the time taken for 50% of the reaction to complete is **16 minutes**.

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