Question

Calculate the two third life of a first reaction having K=5.48 $\times$ 10$^{-14}$s$^{-1}$.

Hint:

Make sure you understand if "2/3 life" means remaining is 2/3 or consumed is 2/3. In kinetics contexts, it usually refers to completion, so remaining = 1 - completion.

Answer 1

"Two-third life" ($t_{2/3}$) is the time required for the reaction to be 2/3 complete.
This means the amount consumed is $2/3$, so the amount remaining $[A]$ is $1 - 2/3 = 1/3$ of $[A]_0$.
So, $\frac{[A]_0}{[A]} = 3$.
Formula: $t = \frac{2.303}{k} \log \frac{[A]_0}{[A]}$
Substitute values ($k = 5.48 \times 10^{-14}$ s$^{-1}$):
$t_{2/3} = \frac{2.303}{5.48 \times 10^{-14}} \log(3)$
We know $\log(3) \approx 0.4771$.
$t_{2/3} = \frac{2.303 \times 0.4771}{5.48 \times 10^{-14}}$
$t_{2/3} = \frac{1.0987}{5.48} \times 10^{14}$
$t_{2/3} \approx 0.2005 \times 10^{14}$
$t_{2/3} \approx 2.0 \times 10^{13}$ s.