The half life of a first order reaction having rate constant, $k = 1.7 × 10^{-5} s^{-1}$ is

11.3 hrs

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

$1 \, \text{hour} = 3600 \, \text{s}$

$t_{1/2} = \frac{40764.7}{3600} \approx 11.3 \, \text{hours}$