Question

Half-life of a first-order reaction is 20 minutes. The time taken to reduce the initial concentration of the reactant to \( \left(\frac{1}{10}\right)^{\text{th}} \) of its initial value is \( \rule{1cm}{0.15mm} \).

• A. 46.60 min
• B. 66.46 min
• C. 79.68 min
• D. 88.00 min

Hint:

For first order reactions: \[ t=\frac{2.303}{k}\log\frac{[A]_0}{[A]} \] and if concentration becomes one-tenth, then \(\log 10 = 1\).

Answer 1

The Correct Option is B

Concept:
For a first order reaction: \[ t_{1/2}=\frac{0.693}{k} \] Also, \[ t=\frac{2.303}{k}\log\frac{[A]_0}{[A]} \] ip

Step 1: Find the rate constant from half-life.
Given: \[ t_{1/2}=20\ \text{min} \] So, \[ k=\frac{0.693}{20}=0.03465\ \text{min}^{-1} \] ip

Step 2: Use the first order formula.
The concentration becomes one-tenth of the initial concentration, so: \[ \frac{[A]_0}{[A]}=10 \] Thus, \[ t=\frac{2.303}{k}\log 10 \] Since \(\log 10 = 1\), \[ t=\frac{2.303}{0.03465} \] ip

Step 3: Calculate the time.
\[ t \approx 66.46\ \text{min} \] ip Hence, the correct answer is:
\[ \boxed{(B)\ 66.46\ \text{min}} \]