The half life of a first order reaction having rate constant, $k = 1.7 × 10^{-5} s^{-1}$ is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 11.3 hrs
We are asked to calculate the half-life of a first-order reaction given the rate constant:
$k = 1.7 \times 10^{-5} \, \text{s}^{-1}$
Step 1: Formula for half-life of a first-order reaction
For a first-order reaction:
$t_{1/2} = \frac{0.693}{k}$
Step 2: Substitute the value of k
$t_{1/2} = \frac{0.693}{1.7 \times 10^{-5}}$
$t_{1/2} = 40764.7 \, \text{s} \quad (\text{approx})$
Step 3: Convert seconds to hours
$1 \, \text{hour} = 3600 \, \text{s}$
$t_{1/2} = \frac{40764.7}{3600} \approx 11.3 \, \text{hours}$