The termolecular reaction
`2NO(g) + H_(2)(g) rarr 2NOH(g)` is found to be third order obeying the rate law `r = k[NO]^(2)[H_(2)]`. Show that it is conisstent with either of the following mechanisms:
(a) `2NO(g) overset(k_(eq))hArr N_(2)O_(2)(g) ("fast equilibrium")`
`N_(2)O_(2)(g) + H_(2)(g) overset(k')rarr 2NOH(g)` (slow)
(b) `2NO(g) +H_(2)(g) overset(k'_(eq))hArr NOH_(2)(g) ("fast equilibrium")`
`NOH_(2)(g) + NO(g) overset(k'')rarr 2NOH(g)` (slow)

Explanation

(a) Since the slow step is the rate-determining step.
Hence, `r = k'[N_(2)O_(2)] [H_(2)]`
`rArr r = k' k_(eq) [NO]^(2)[H_(2)] = k[NO]^(2)[H_(2)]` (where `k = k' k_(eq)`)
(b) Proceeding as in (a) above, we have
`r = k''[NOH_(2)] [NO]`
and form the fast equilibrium step,
`k_(eq) = ([N_(2)O_(2)])/([NO]^(2))` or `[N_(2)O_(2)] = k_(aq)[NO]^(2)`
and form the fast equilibrium step,
`k'_(eq) = ([NOH_(2)])/([NO][H_(2)])` or `[NOH]_(2) = k'_(eq) [NO][H_(2)]`
`rArr r = k'' k'_(eq) [NO]^(2)[H_(2)] = k[NO]^(2)[H_(2)]` (where `k = k'' k'_(eq)`)