Di-tert-butyl peroxide vapour breaks down when it is heated…

The pressure of the peroxide. Let the peroxide fall from $p_0$ to $p$. Then acetone is $2(p_0 - p)$ and ethane is $(p_0 - p)$, so the total is $p_t = p + 3(p_0 - p) = 3p_0 - 2p$, and $p = \dfrac{3p_0 - p_t}{2}$. At $25.0\ \text{min}$: $p = \dfrac{0.900 - 0.546}{2} = 0.1770\ \text{bar}$.

(a) Rate constant. $k = \dfrac{1}{25.0}\ln\dfrac{0.300}{0.1770} = \dfrac{\ln 1.6949}{25.0}\ \text{min}^{-1} = \dfrac{0.5276}{25.0}\ \text{min}^{-1} = 0.0211\ \text{min}^{-1}$. The second reading agrees: at $50.0\ \text{min}$, $p = 0.1045\ \text{bar}$, which gives $k = 0.0211\ \text{min}^{-1}$. A constant $k$ is the sign of a first order reaction.

(b) Half-life. $t_{1/2} = \dfrac{0.693}{k} = 32.8\ \text{min}$.

(c) Acetone. $2(p_0 - p) = 2(0.300 - 0.1770)\ \text{bar} = 0.246\ \text{bar}$.

(d) Equal pressures. $2(p_0 - p) = p$ gives $p = \tfrac{2}{3}p_0 = 0.200\ \text{bar}$. Then $t = \dfrac{1}{k}\ln\dfrac{p_0}{p} = \dfrac{\ln 1.5}{0.0211}\ \text{min} = 19.2\ \text{min}$.

(e) After $80.0\ \text{min}$. $p = 0.300\,e^{-1.688}\ \text{bar} = 0.0554\ \text{bar}$, so $p_t = 0.900 - 2 \times 0.0554 = 0.789\ \text{bar}$.