Question

For a two step reaction.
`AhArrR+B" "R+Coverset(k_(2))toP`
(where, R is a reactive intermediate whose concentration is maintained at some low steady state throughout the reaction). If the concentration of C is very high then the order of reaction for formation of "P" is
A. 2
B. 0
C. 1
D. `1//2`

Answer 1

Correct Answer - C
Rate law `=(dp)/(dt)=K_(2)[R][C]`
at steady state `(d[R])/(dt)=0`
`(d[R])/(dt)=K_(1)[A]-K_(2)[R][B]-K_(2)[R][C]=0`
`K_(1)[A]={K_(2)[B]+K_(2)[C]}[R]`
`[R]=(K_(1)[A])/(K_(2){[B]+[C]})`
Rate `=K_(2)xx(K_(1))/(K_(2))([A])/([B]+[C])xx[C]`
`ul(K_(1)[A])`
`=([B])/([C])+1`
where C is very high
Rate `=K_(1)[A]`
`:.` order of reaction` = 1 " "]`