Answer
$t_{1/2} = 75\text{ s}$
Work Step by Step
First, take note of the formula for first-order reactions given by
ln$\bigg ( \dfrac{[A]}{[A]_0}\bigg ) = -kt$
Then take note of the values given for [A]$_0$ = 100.0 which after 65 seconds loses 45% as it reacts becoming [A] = 55.0.
We use this to solve for the constant k in the reaction,
ln$\bigg ( \dfrac{[55.0]}{[100.0]}\bigg ) = -k(65 \ s)$, k = $9.2 \times 10^{-3} \ s^{-1}$
Thus the half-life for the compound is,
t$_{1/2}=\dfrac{ln \ 2}{ \ k} = \dfrac{0.693}{9.2 \times 10^{-3} \ s^{-1}} = 75 \ s$
