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Q.

The reaction:

OCl+I- OH-OI+Cl

takes place in the following steps:

i. OCl+H2Ok1k2HOCl+OH        (fast)

ii. I+HOClk3HOI+Cl            (slow)

iii. OH+HOIk1k2H2O+OI        (fast)

The rate of consumption of I in the following equation is

 

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a

k3k1k2OClI[OH]2

b

k1k2k3OClI[OH]

c

k1k3k2OClI[OH]

d

k2k3k1OClI[OH]

answer is C.

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Detailed Solution

detailed_solution_thumbnail

Step (ii) is the rate-determining step.

 Rate =k3I[HOCl]

(HOCI) produced in step (i) is consumed in step (ii), so it is a reactive intermediate. Similarly, (HOI) produced in step (ii) is consumed in step (iii), so it is also a reactive intermediate. The concentration of both reactive species is determined from
their equilibrium constant value.

 from step (i)

k1k2=[HOCl][OH]OCl[HOCl]=k1k2OCl[OH]   .. . (2) 

Substitute the value of [HOCI] from Eq. (2) in Eq. (1),

 Rate =k1k3k2OClI[OH]

Hence answer is (3). 

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