Question

The half life period of a first order reaction is 1 min 40 sec. Calculate its rate constant.

• A. \(6.93\times 10^{-3}\) min\(^{-1}\)
• B. \(6.93\times 10^{-3}\) sec\(^{-1}\)
• C. \(6.93\times 10^{-3}\) sec
• D. \(6.93\times 10^{-3}\) min

Hint:

For first order: \(t_{1/2}=\frac{0.693}{k}\). Always keep units consistent with time.

Answer 1

The Correct Option is B

Step 1: Convert half-life into seconds.
\[ t_{1/2} = 1 \text{ min } 40 \text{ sec } = 60 + 40 = 100 \text{ sec} \]
Step 2: Use first order half-life relation.
\[ t_{1/2} = \frac{0.693}{k} \Rightarrow k = \frac{0.693}{t_{1/2}} \]
Step 3: Substitute value.
\[ k = \frac{0.693}{100} = 6.93 \times 10^{-3}\, s^{-1} \]
Final Answer:
\[ \boxed{6.93\times 10^{-3}\,s^{-1}} \]