Question

Rate constant for first order reaction is \(40s^{-1}\). Calculate the time required to reduce the concentration to \(1/20\) th of its initial value.

• A. \(3.50\times 10^{-2}\) s
• B. \(5.12\times 10^{-2}\) s
• C. \(4.50\times 10^{-2}\) s
• D. \(7.48\times 10^{-2}\) s

Hint:

Use the integrated rate law of a first order reaction with the final concentration equal to one twentieth of the initial value.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
For a first order reaction the time taken is given by \(t=\dfrac{2.303}{k}\log\dfrac{[A]_0}{[A]}\).

Step 2: Substitute the Data
Here \(k=40\ \text{s}^{-1}\) and \([A]=\dfrac{[A]_0}{20}\), so \(\dfrac{[A]_0}{[A]}=20\).

Step 3: Calculation
\(t=\dfrac{2.303}{40}\log 20=\dfrac{2.303\times 1.30}{40}\)
\(t\approx 7.48\times 10^{-2}\ \text{s}\)

Step 4: Compare the options
This value matches the option \(7.48\times 10^{-2}\) s.

Final Answer:
The time required is about 0.0748 s. \[ \boxed{\text{(D) }7.48\times 10^{-2}\text{ s}} \]