Chemistry - Chemical Kinetics Question with Solution | TestHub

$\begin{array}{cccccc}\text{Step I}& 2\text{NO}& \stackrel{{\text{k}}_1}{\rightleftharpoons }& {\text{N}}_2{\text{O}}_2& \text{;}& {\text{K}}_{\text{eq}}\left(\text{fast}\right)\\ \text{Step II}& {\text{N}}_2{\text{O}}_2+{\text{H}}_2& \stackrel{{\text{k}}_2}{⟶}& {\text{N}}_2\text{O}+{\text{H}}_2\text{O}& & \left(\text{slow}\right)\\ \text{Step III}& {\text{N}}_2\text{O}+{\text{H}}_2& \stackrel{{\text{k}}_3}{⟶}& {\text{N}}_2+{\text{H}}_2\text{O}& & \left(\text{fast}\right)\end{array}$

(i)  Derive the rate Law

$\text{Rate}\text{=}{\text{k}}_{\text{2}}\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]\left[{\text{H}}_{\text{2}}\right]$(Rate determining step)

$\frac{\left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]}{{\left[\text{NO}\right]}^{\text{2}}}\text{=}{\text{K}}_{\text{eq}}\therefore \left[{\text{N}}_{\text{2}}{\text{O}}_{\text{2}}\right]\text{=}{\text{K}}_{\text{eq}}{\left[\text{NO}\right]}^{\text{2}}$

$\text{Rate}\text{=}{\text{k}}_{\text{2}}{\text{K}}_{\text{eq}}{\left[\text{NO}\right]}^{\text{2}}\left[{\text{H}}_{\text{2}}\right]$

k'${\left[\text{NO}\right]}^{\text{2}}\left[{\text{H}}_{\text{2}}\right]$Rate law is third order