Question

The two-third life (\(t_{2/3}\)) of a first order reaction in which \(k = 5.48 \times 10^{-14} \, \text{per sec}\), is

• A. \(\frac{2.303}{5.48 \times 10^{-14}} \log 3\)
• B. \(\frac{2.303}{5.48 \times 10^{-14}} \log 2\)
• C. \(\frac{2.303}{5.48 \times 10^{-14}} \log \frac{1}{3}\)
• D. \(\frac{2.303}{5.48 \times 10^{-14}} \log \frac{2}{3}\)

Hint:

For a first-order reaction, \(t_{1/2} = \frac{0.693}{k}\), \(t_{2/3} = \frac{2.303}{k} \log 3\), and \(t_{3/4} = \frac{2.303}{k} \log 4 = \frac{0.693}{k} \times 2 = 2t_{1/2}\).

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
For a first-order reaction, the integrated rate law is \(k = \frac{2.303}{t} \log \frac{[A]_0}{[A]}\).
Step 2: Detailed Explanation:
\(t_{2/3}\) is the time when two-thirds of the reaction is complete, i.e., one-third remains. So, \(\frac{[A]}{[A]_0} = \frac{1}{3}\). \[ k = \frac{2.303}{t_{2/3}} \log \frac{[A]_0}{[A]} = \frac{2.303}{t_{2/3}} \log 3 \] \[ t_{2/3} = \frac{2.303}{k} \log 3 \] Substituting \(k = 5.48 \times 10^{-14}\), \[ t_{2/3} = \frac{2.303}{5.48 \times 10^{-14}} \log 3 \]
Step 3: Final Answer:
The correct expression is option (A).