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The reaction: OCl^(ɵ) + I^(ɵ)overset(...

The reaction:
`OCl^(ɵ) + I^(ɵ)overset(overset(ɵ)(OH))rarr OI^(ɵ)+Cl^(ɵ)`
takes place in the following steps:
(i) `OCl^(ɵ)+H_(2)O underset(k_(2))overset(k_(1))hArr HOCl+overset(ɵ) (OH) ("fast")`
(ii) `I^(ɵ) + HOCloverset(k_(3))rarrHOI + Cl^(ɵ)` (slow)
(iii) `overset(ɵ)(OH) + HOI underset(k_(1)')overset(k_(2)')hArr H_(2)O+OI^(ɵ)" " ("fast")`
The rate of consumption of `I^(ɵ)` in the following equation is

A

`(k_(3)k_(1))/(k_(2))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)]^(2))`

B

`(k_(1)k_(2))/(k_(3))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)])`

C

`(k_(1)k_(3))/(k_(2))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)])`

D

`(k_(2)k_(3))/(k_(1))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)])`

Text Solution

Verified by Experts

The correct Answer is:
C
Step (ii) is the rate-determineing step.
`:.` Rate `= k_(3)[I^(ɵ)][HOCl]` …(i)
`(HOCl)` profuced in step (i) is consumed in step (ii), so it is a reactive intermediate. ismilarly, `(HOI)` profuced in step (ii) is consumed in step (iii) ,so it is also a reactive intermediate. The concentration of both reactive species is determined form their equilibrium constant value.
`:.` form step (i)
`(k_(1))/(k_(2)) = ([HOCl][overset(ɵ)(OH)])/([OCl^(ɵ)]) rArr [HOCl] = (k_(1))/(k_(2))([overset(ɵ)(OCl)])/([overset(ɵ)(OH)])` ....(2)
Substituting the value of `[HOCl]` form Eq. (2) in Eq. (1) ,
`Rate = (k_(1)k_(3))/(k_(2))([OCl^(ɵ)][I^(ɵ)])/([overset(ɵ)(OH)])`
Hence answer is (c ).
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