The rate law for the decompoistion of gaseous `N_(2)O_(5)`,
`N_(2)O_(5)(g) rarr 2NO_(2)(g) + (1)/(2)O_(2)(g)` is observed to be:
`r = (-d[N_(2)O_(5)])/(dt) = k[N_(2)O_(5)]`
A reaction machanism which has been suggested to be conisstent with this rate law is
`N_(2)O_(5)(g) overset(k_(eq))hArr NO_(2)(g)+NO_(3)(g) ("fast equilibrium")`
`NO_(2) (g) + NO_(3)(g) overset(k_(1))rarr NO_(2)(g) + NO(g) + O_(2)(g)` (slow)
`NO(g) + NO_(3)(g) overset(k_(2))rarr 2NO_(2)(g)` (fast)`
Show that the mechanism is consistent with the above rate law.

Explanation

Since the slow step is the rate determining step, hence
`r=k_(1)[NO_(2)][NO_(3)]" "(1)` and from the fast equilibrium step,
`K=([NO_(2)][NO_(3)])/([N_(2)O_(5)])`
Thus, `[NO_(2)][NO_(3)]=K[N_(2)O_(5)]" "(ii)`
Using (ii) in (i), we get :
`r=k_(1)K[N_(2)O_(5)]=k[N_(2)O_(5)]`
where `k=k_(1)K`
This shows that the mechanism is consistent with the observed rate law.