Here are the solutions to the questions: 1. (d) (i) Determine the overall order of the reaction $2\text{NO(g)} + \text{Cl}_2\text{(g)} \to 2\text{NOCl(g)}$ showing all the steps. The general rate law for this reaction can be written as: $$ \text{Rate} = k[\text{NO}]^x[\text{Cl}_2]^y $$ where $x$ is the order with respect to $\text{NO}$, $y$ is the order with respect to $\text{Cl}_2$, and $k$ is the rate constant. Step 1: Determine the order with respect to $[\text{Cl}_2]$ ($y$). Compare Experiment 1 and Experiment 2, where $[\text{NO}]$ is kept constant. From Experiment 1: $\text{Rate}_1 = k(0.10)^x(0.10)^y = 0.17$ From Experiment 2: $\text{Rate}_2 = k(0.10)^x(0.20)^y = 0.35$ Divide $\text{Rate}_2$ by $\text{Rate}_1$: $$ \frac{\text{Rate}_2}{\text{Rate}_1} = \frac{k(0.10)^x(0.20)^y}{k(0.10)^x(0.10)^y} = \frac{0.35}{0.17} $$ $$ \left(\frac{0.20}{0.10}\right)^y = 2.0588 $$ $$ (2)^y \approx 2 $$ Therefore, $y = 1$. The reaction is first order with respect to $[\text{Cl}_2]$. Step 2: Determine the order with respect to $[\text{NO}]$ ($x$). Compare Experiment 2 and Experiment 3, where $[\text{Cl}_2]$ is kept constant. From Experiment 2: $\text{Rate}_2 = k(0.10)^x(0.20)^y = 0.35$ From Experiment 3: $\text{Rate}_3 = k(0.20)^x(0.20)^y = 1.45$ Divide $\text{Rate}_3$ by $\text{Rate}_2$: $$ \frac{\text{Rate}_3}{\text{Rate}_2} = \frac{k(0.20)^x(0.20)^y}{k(0.10)^x(0.20)^y} = \frac{1.45}{0.35} $$ $$ \left