Question

When the value of rate constant \((k)\) is \(2.0 \, \mathrm{min^{-1}}\), then what will be half time of reaction \((t_{1/2})\) in seconds?

• A. 34.4
• B. 24.6
• C. 30.2
• D. 20.8

Hint:

For a first-order reaction, always use \(t_{1/2} = \frac{0.693}{k}\). If the answer is asked in seconds, first calculate using the given unit and then convert carefully.

Answer 1

The Correct Option is D

Step 1: Identify the order of reaction from the unit of rate constant.
The unit of the rate constant is given as \(\mathrm{min^{-1}}\). This unit corresponds to a first-order reaction. For a first-order reaction, the half-life is independent of the initial concentration and is given by a fixed formula.
Step 2: Write the formula for half-life of first-order reaction.
For a first-order reaction,
\[ t_{1/2} = \frac{0.693}{k} \] Here,
\[ k = 2.0 \, \mathrm{min^{-1}} \] So,
\[ t_{1/2} = \frac{0.693}{2.0} = 0.3465 \, \mathrm{min} \] Step 3: Convert the time into seconds.
Since,
\[ 1 \, \mathrm{min} = 60 \, \mathrm{s} \] therefore,
\[ t_{1/2} = 0.3465 \times 60 = 20.79 \, \mathrm{s} \] \[ t_{1/2} \approx 20.8 \, \mathrm{s} \] Step 4: Conclusion.
Hence, the half-life of the reaction is \(20.8\) seconds. Therefore, the correct option is \((D)\).
Final Answer:20.8 s.