For the reaction mechanism of the reaction.
`2 NO (g) + 2 H _(2) (g) to N _(2) (g) + 2 H _(2) O (g).`
`{:("Step l", 2 NO , overset(k _(1)) hArr, N _(2) O _(2):, K _(eq) ("fast")), ("Step ll", N _(2) O _(2) + H _(2), overset( k_(2)) to , N _(2) O + H _(2) O , ("slow")), ("Step lll", N _(2) O + H _(2), overset( K _(3)) to ,N _(2) + H _(2) O , ("fast")):}`
Expresson of rate of reaction is
(Take `k _(eq) xx k _(2) = k`)

Options

A. `k ' N_(2) O _(2) [ H _(2)]`
B. `k'N _(2) O [H_(2)]`
C. `k'N _(2) O[H_(2)]`
D. `k ' N _(2) O _(2)`

Answer

A

Explanation

`{:("Step l", 2 NO , overset(k _(1)) hArr, N _(2) O _(2):, K _(eq) ("fast")), ("Step ll", N _(2) O _(2) + H _(2), overset( k_(2)) to , N _(2) O + H _(2) O , ("slow")), ("Step lll", N _(2) O + H _(2), overset( K _(3)) to ,N _(2) + H _(2) O , ("fast")):}`
(l) Derive the rate Law
Rate `k _(2) [ N _(2) O _(2) ] [ H _(2) ]` (Rate determining step)
`([ N _(2) O _(2)])/( [NO ] ^(2)) = k _(eq) therefore [ N _(2) O _(2) ] = K _(eq) [ NO ] ^(2)`
Rate = `k _(2) k _(eq) [ NO ] ^(2) [ H _(2)]`
`= k. [ NO ] ^(2) [ H _(2) ]` Hence, the reaction is third order.