$2 \mathrm{NO}_{(g)}+\mathrm{Cl}_{2(g)} ightleftharpoons 2 \mathrm{NOCl}_{(s)}$ \This reaction wa

Question

$2 \mathrm{NO}_{(g)}+\mathrm{Cl}_{2(g)} ightleftharpoons 2 \mathrm{NOCl}_{(s)}$ \This reaction was studied at $-10{ }^{\circ} \mathrm{C}$ and the following data was obtained. $$\begin{tabular}{|c|c|c|c|} \hline Run & $[\mathrm{NO}]_{\mathbf{0}}$ & $\left[\mathrm{Cl}_{2}ight]_{\mathbf{0}}$ & $r_{0}$ \ \hline 1 & 0.10 & 0.10 & 0.18 \ \hline 2 & 0.10 & 0.20 & 0.35 \ \hline 3 & 0.20 & 0.20 & 1.40 \ \hline \end{tabular}$$ $[\mathrm{NO}]_{0}$ and $\left[\mathrm{Cl}_{2}ight]_{0}$ are the initial concentrations and $r_{0}$ is the initial reaction rate.\ The overall order of the reaction is _______.\ (Nearest Integer).

Professor PC Thomas Classes · Answer

(3) : $R=k[\mathrm{NO}]^{x}\left[\mathrm{Cl}_{2}ight]^{y}$\ $0.18=k[0.1]^{x}[0.1]^{y}$\ $0.35=k[0.1]^{x}[0.2]^{y}$\ $1.40=k[0.2]^{x}[0.2]^{y}$\ On dividing (ii) by (i)\ $\frac{0.35}{0.18}=\frac{k[0.1]^{x}[0.2]^{y}}{k[0.1]^{x}[0.1]^{y}}$\ $\frac{0.35}{0.18}=\left(\frac{0.2}{0.1}ight)^{y} \quad$ or $2^{y}=2^{1} \quad$ or $y=1$\ On dividing (iii) by (ii) $$ \begin{aligned} & \frac{1.40}{0.35}=\frac{k[0.2]^{x}[0.2]^{y}}{k[0.1]^{x}[0.2]^{y}} \ & 4=2^{x} 	ext { or } 2^{2}=2^{x} \Rightarrow x=2 \end{aligned} $$ So, the rate law expression\ $\Rightarrow \quad R=k[\mathrm{NO}]^{2}\left[\mathrm{Cl}_{2}ight]^{1}$\ The overall order of reaction $=2+1=3$\