Question

Half life period of a first order reaction is \(1386\) seconds. The rate constant of the reaction is

• A. \(5.5\times 10^{-2}\times s^{-1}\)
• B. \(0.5\times 10^{-3}\times s^{-1}\)
• C. \(5\times 10^{-2}\times s^{-1}\)
• D. \(5\times 10^{-3}\times s^{-1}\)

Hint:

For a first order reaction, k = 0.693 / t half.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
For a first-order reaction the half-life does not depend on the initial concentration. It is linked to the rate constant by a fixed relation.

Step 2: Key Formula:
\[ k = \dfrac{0.693}{t_{1/2}} \]

Step 3: Substitute:
\[ k = \dfrac{0.693}{1386} = 5 \times 10^{-4} \text{ s}^{-1} = 0.5 \times 10^{-3} \text{ s}^{-1} \]
This follows since \(1386 = 2 \times 693\), so \(\dfrac{693 \times 10^{-3}}{1386} = 0.5 \times 10^{-3}\).

Step 4: Why the other options are wrong.
\(5.5 \times 10^{-2}\), \(5 \times 10^{-2}\) and \(5 \times 10^{-3}\) s\(^{-1}\) are 110, 100 and 10 times too large, which would give half-lives of about 12.6 s, 13.9 s and 139 s.

Final Answer:
The rate constant is \(0.5 \times 10^{-3}\) s\(^{-1}\). \[ \boxed{\text{(B) }0.5 \times 10^{-3}\ \text{s}^{-1}} \]