Consider this reaction,

$2{\mathrm{H}}_2\left(g\right)+2\mathrm{NO}\left(g\right)⟶{\mathrm{N}}_2\left(g\right)+2{\mathrm{H}}_2\mathrm{O}\left(g\right)$

The rate law for this reaction is rate $=k\left[{\mathrm{H}}_2\right][\mathrm{NO}{]}^2$

Under what conditions could these steps represent the mechanism?

$\text{ Step 1 : 2NO }\rightleftharpoons {\mathrm{N}}_2{\mathrm{O}}_2$

$\text{ Step 2: }{\mathrm{N}}_2{\mathrm{O}}_2+{\mathrm{H}}_2⟶{\mathrm{N}}_2\mathrm{O}+{\mathrm{H}}_2\mathrm{O}$

$\text{ Step 3: }{\mathrm{N}}_2\mathrm{O}+{\mathrm{H}}_2⟶{\mathrm{N}}_2+{\mathrm{H}}_2\mathrm{O}$

The response is as follows:

$2{\mathrm{H}}_2\left(\mathrm{g}\right)+2\mathrm{NO}\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)$

The rate law is as follows:

$\text{ Rate }=\mathrm{K}\left[{\mathrm{H}}_2\right][\mathrm{NO}{]}^2$

The slowest stage determines the rate of the chemical reaction. Therefore, since the rate of the reaction is dictated by that, we should have 2 molecules of $NO$ and 1 molecule of ${\mathrm{H}}_2$ in the slowest step.

$\text{ So, I. }2\mathrm{NO}\left(\mathrm{g}\right)+{\mathrm{H}}_2\left(\mathrm{g}\right)\to {\mathrm{N}}_2\left(\mathrm{g}\right)+{\mathrm{H}}_2{\mathrm{O}}_2\text{ (slow) }$

$\text{ II. }{\mathrm{H}}_2{\mathrm{O}}_2+{\mathrm{H}}_2\left(\mathrm{g}\right)\to 2{\mathrm{H}}_2\mathrm{O}\left(\mathrm{g}\right)\text{ (fast) }$

This might be the reaction's mechanism, according to rate law.