Chemical Kinetics Part 1

According to given reaction,
$2\mathit{N}\mathit{O}\left(\mathit{g}\right)+{\mathit{C}\mathit{l}}_2\left(\mathit{g}\right)\rightleftharpoons 2\mathit{N}\mathit{O}\mathit{C}\mathit{l}\left(\mathit{s}\right)$
Given data are as follows,
By rate law expression, $\mathit{r}=\mathit{k}[{\mathit{N}\mathit{O}}^{\mathit{x}}{\left[{\mathit{C}\mathit{l}}_2\right]}^{\mathit{y}}$
$\mathit{E}\mathit{x}\mathit{p}$ (I) $0.18=\mathit{k}{\left(0.1\right)}^{\mathit{x}}{\left(0.1\right)}^{\mathit{y}}$ ...(i)
$\mathit{E}\mathit{x}\mathit{p}$ (II) $0.35=\mathit{k}{\left(0.1\right)}^{\mathit{x}}{\left(0.2\right)}^{\mathit{y}}$...(ii)
$\mathit{E}\mathit{x}\mathit{p}$ (III) $1.40=\mathit{k}{\left(0.2\right)}^{\mathit{x}}{\left(0.2\right)}^{\mathit{y}}$...(iii)
Dividing equation (ii) and (iii), we get
$\frac{0.35}{1.40}=\frac{\mathit{k}{\left(0.1\right)}^{\mathit{x}}{\left(0.2\right)}^{\mathit{y}}}{\mathit{k}{\left(0.2\right)}^{\mathit{x}}{\left(0.2\right)}^{\mathit{y}}}$
$\frac{1}{4}={\left(\frac{1}{2}\right)}^{\mathit{x}}\Rightarrow \mathit{x}=2$
Similarly dividing equation (i) and (ii), we get
$\frac{1}{2}={\left(\frac{1}{2}\right)}^{\mathit{y}}\Rightarrow \mathit{y}=1$
Now, $\mathit{x}+\mathit{y}=$ order of reactions
$\mathit{x}+\mathit{y}\Rightarrow 2+1\Rightarrow 3$
Overall, order of the reaction is 3 .