Chemistry - Chemical Kinetics Question with Solution | TestHub

${\text{K}}_{\text{eq}}=\frac{{\text{k}}_{\text{f}}}{{\text{k}}_{\text{b}}}=\frac{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\text{O}\right]}^2}{{\left[{\text{H}}_2\right]}^2{[\text{NO}]}^2}$

At equilibrium${\text{r}}_{\text{f}}={\text{r}}_{\text{b}}$

${\text{k}}_{\text{f}}\left[{\text{H}}_2\right]{[\text{NO}]}^2={\text{k}}_{\text{b}}\frac{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\text{O}\right]}^2}{\left[{\text{H}}_2\right]}$

Hence, rate expression for reverse reaction.

$={\text{k}}_{\text{b}}\frac{\left[{\text{N}}_2\right]{\left[{\text{H}}_2\text{O}\right]}^2}{\left[{\text{H}}_2\right]}$