The half-life of a first-order reaction having rate constant $k = 1.7 \times 10^{- 5} {s e c}^{- 1}$ is:

12.1 h r s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

9.7 h r s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

11.3 h r s

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

1.8 h r s

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C $11.3 h r s$
$t_{1 / 2} = \frac{l n 2}{k} = \frac{0.693}{1.7 \times 10^{- 5}} = 40764 s e c = \frac{40764}{3600} h r s = 11.3 h r s .$

Suggest Corrections

Similar questions

k = 0.231 \times 10^{- 10} s^{- 1}

K = 5.5 \times 10^{- 14} S^{- 1}

5.5 \times 10^{- 14} {s e c}^{- 1}

(a) Derive integrated rate equation for $1^{s t}$ order rate equation

(b) A first order reaction is found to have a rate constant, $k = 5.5 \times 10^{- 14} S^{- 1}$ Find the half-life of the reaction.

1.54 \times 10^{- 3} s e c^{- 1}

Join BYJU'S Learning Program