(i) Determine the overall order of the reaction 2NO(g) + Cl₂(g) 2NOCl(g) showing all the steps.

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 $\text{Rate}_2$:

$$\frac{\text{Rate}_3}{\text{Rate}_2} = \frac{k(0.20)^x(0.20)^y}{k(0.10)^x(0.20)^y} = \frac{1.45}{0.35}$$ $$$$