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 of determining when 50% of a first-order reaction was completed, given that 75% of the reaction was completed in 32 minutes, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the First-Order Reaction Kinetics**:
For a first-order reaction, the relationship between the rate constant (k) and time (t) can be expressed using the formula:
\[ k = \frac{2.303}{t} \log \left( \frac{[A]_0}{[A]_t} \right) \]
where \([A]_0\) is the initial concentration and \([A]_t\) is the concentration at time \(t\).

2. **Set Up the First Scenario (75% Completion)**:
- Let’s assume the initial concentration \([A]_0 = 100\) (arbitrary units).
- If 75% of the reaction is completed, then the concentration at time \(t\) (after 32 minutes) is:
\[ [A]_t = [A]_0 - 75\% \cdot [A]_0 = 100 - 75 = 25 \]
- Now, substitute these values into the first-order reaction formula:
\[ k = \frac{2.303}{32} \log \left( \frac{100}{25} \right) \]
- Simplifying the logarithm:
\[ \log \left( \frac{100}{25} \right) = \log(4) \]

3. **Set Up the Second Scenario (50% Completion)**:
- For 50% completion, the concentration at time \(t\) is:
\[ [A]_t = [A]_0 - 50\% \cdot [A]_0 = 100 - 50 = 50 \]
- Substitute these values into the first-order reaction formula:
\[ k = \frac{2.303}{x} \log \left( \frac{100}{50} \right) \]
- Simplifying the logarithm:
\[ \log \left( \frac{100}{50} \right) = \log(2) \]

4. **Equate the Two Scenarios**:
Since the rate constant \(k\) is the same for both scenarios, we can set the two equations equal to each other:
\[ \frac{2.303}{32} \log(4) = \frac{2.303}{x} \log(2) \]

5. **Cancel Out Common Factors**:
- The factor \(2.303\) cancels out:
\[ \frac{1}{32} \log(4) = \frac{1}{x} \log(2) \]

6. **Substitute Logarithm Values**:
- We know that \(\log(4) = \log(2^2) = 2 \log(2)\):
\[ \frac{1}{32} \cdot 2 \log(2) = \frac{1}{x} \log(2) \]
- Now, we can cancel \(\log(2)\) from both sides (assuming \(\log(2) \neq 0\)):
\[ \frac{2}{32} = \frac{1}{x} \]

7. **Solve for \(x\)**:
- Rearranging gives:
\[ x = \frac{32}{2} = 16 \text{ minutes} \]

### Final Answer:
50% of the reaction was completed in **16 minutes**.