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: Time in seconds:
1 minute 40 seconds is 60 s plus 40 s, which is 100 s.

Step 2: Use ln 2:
$k = \ln 2 / t_{1/2} = 0.6931/100$.

Step 3: Result:
$k = 6.93 \times 10^{-3}\ \text{s}^{-1}$. A nice round half-life like 100 s gives a clean value, which is a good sign the conversion is right.

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