Question

What is the time required for 75 % completion of a first order reaction if rate constant is \(23.03 \text{minute}^{-1}\) ?

• A. \(12.00\) s
• B. \(3.6\) s
• C. \(36\) s
• D. \(6.0\) s

Hint:

Use t = (2.303/k) log(100/25) and watch that k is per minute while options are in seconds.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a first order reaction, the time to reach a given fraction of completion depends only on \(k\). Here 75 % complete means 25 % of the reactant is left.

Step 2: Key Formula or Approach:
\[ t = \frac{2.303}{k}\log\frac{[A]_0}{[A]} \]
If 75 % reacts, \([A]_0/[A] = 100/25 = 4\).

Step 3: Detailed Explanation:
\[ t = \frac{2.303}{23.03 \text{ min}^{-1}}\log 4 = 0.1 \text{ min} \times 0.6021 = 0.06021 \text{ min} \]
Convert minutes to seconds:
\[ t = 0.06021 \times 60 = 3.61 \text{ s} \approx 3.6 \text{ s} \]
Option (C) 36 s comes from forgetting the factor 0.1 in \(2.303/23.03\). Options (A) and (D) arise from using the wrong time unit or using 50 percent completion, which would give 1.8 s.

Final Answer:
The time required for 75 percent completion is about 3.6 s, option (B). \[ \boxed{3.6 \text{ s (B)}} \]