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

Step 1: Understanding the Concept:
For a first-order reaction, the time required for a certain fraction of the reaction to complete is related to the rate constant (\(k\)) and initial/final concentrations.
Step 2: Key Formula or Approach:
\[ k = \frac{0.693}{t_{1/2}} \quad \text{and} \quad t = \frac{2.303}{k} \log \left( \frac{[A]_0}{[A]_t} \right) \] Step 3: Detailed Explanation:
Given:
\(t_{1/2} = 20 \text{ min}\)
\([A]_t = \frac{1}{10} [A]_0 \rightarrow \frac{[A]_0}{[A]_t} = 10\)
1. Calculate \(k\):
\[ k = \frac{0.693}{20} = 0.03465 \text{ min}^{-1} \] 2. Calculate time \(t\):
\[ t = \frac{2.303}{0.03465} \log(10) \] Since \(\log(10) = 1\):
\[ t = \frac{2.303 \times 20}{0.693} \] \[ t \approx 3.322 \times 20 = 66.44 \text{ min} \] (Rounding gives 66.46 min from options).
Step 4: Final Answer:
The time taken is 66.46 min.