Question

Calculate rate constant of a first order reaction having half life \(1\) minute \(40\) second?

• A. \(1.76\times 10^{-3} \text{s}^{-1}\)
• B. \(2.31\times 10^{-3} \text{s}^{-1}\)
• C. \(6.93\times 10^{-3} \text{s}^{-1}\)
• D. \(4.61\times 10^{-3} \text{s}^{-1}\)

Hint:

Convert the half-life to seconds, then k = 0.693 / t half.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
For a first-order reaction the rate constant and the half-life are linked by a fixed relation.

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

Step 3: Detailed Explanation:
Convert the half-life into seconds first: 1 min 40 s \(= 60 + 40 = 100\) s.
\[ k = \frac{0.693}{100\ \text{s}} = 6.93 \times 10^{-3}\ \text{s}^{-1} \]

Step 4: Why the other options are wrong.
\(2.31\times10^{-3}\) would come from \(t_{1/2} = 300\) s, and \(1.76\times10^{-3}\) and \(4.61\times10^{-3}\) from other wrong conversions, such as using 1.40 minutes or 150 s.

Final Answer:
The rate constant is \(6.93\times10^{-3}\ \text{s}^{-1}\), option (C). \[ \boxed{6.93\times10^{-3}\ \text{s}^{-1}} \]