t_ 100\% is the time required for the 100\% completion of the...

For a zero-order reaction:
The integrated rate law is given by $t = \dfrac{[A]_0 - [A]_t}{k}$.

For $50\%$ completion ($t_{1/2}$), $[A]_t = \dfrac{[A]_0}{2}$:
$t_{1/2} = \dfrac{[A]_0 - \dfrac{[A]_0}{2}}{k} = \dfrac{[A]_0}{2k}$

For $100\%$ completion ($t_{100\%}$), $[A]_t = 0$:
$t_{100\%} = \dfrac{[A]_0 - 0}{k} = \dfrac{[A]_0}{k}$

Comparing the two, we get:
$t_{100\%} = 2t_{1/2}$

For a first-order reaction:
The integrated rate law is given by $t = \dfrac{1}{k} \ln\left(\dfrac{[A]_0}{[A]_t}\right)$.

For $100\%$ completion ($t_{100\%}$), $[A]_t = 0$:
$t_{100\%} = \dfrac{1}{k} \ln\left(\dfrac{[A]_0}{0}\right) = \infty$

A first-order reaction takes infinite time for $100\%$ completion. Among the given choices, this is symbolically represented as $\infty$ or $(t_{1/2})^\infty$.

Thus, the correct relation is $t_{100\%} = 2t_{1/2}$ for zero-order and $t_{100\%} = \infty$ for first-order.

Answer: $t_{100\%} = 2t_{1/2}$ and $t_{100\%} = (t_{1/2})^\infty$